I can solve this problem when I put x=2sina, y=2cosa then equation is satisfied. And I put value of x and y in last equation like 6sina+8cosa then Max value is √6^2+8^2=10 . I wait for you your reply. I am a jee student from India🇮🇳
Since x, y lies on the circle with radius 2 we can write x=2cost,y=2sint now 3x +4y becomes 6cost+8sint and we know that max. value of acosx+bsinx is √(a^2+b^2) so the max. value of our expression will be ✓(6^2+8^2)=10
I used a similar method but I didn't use trigonometry, just equations of lines: x²+y²=4 is the equation of the circle centered on O of coordinates (0,0) and of radius 2. 3x+4y=K, where K is a real number, is the equation of a line. We're looking at the maximum value of K where the line intersects the circle. We can see that when K rises, our lines moves upwards. If K is too low or too high, there are no points of the line that are also on the circle and when K rises inside the range where the line and the circle meet each other, the common points move to the north-east. So the value of K that we're looking for is where the line is tangent to the circle in the north-east. But since the equation of the line is 3x+4y=K and since the radius is coming from O, we know that the radius belongs to the line of equation: y=4/3.x. We report this into the equation of the circle: 25/9.x²=4 => x²=36/25 => x=6/5 (x is positive). y=4/3.(6/5)=8/5. Then 3x+4y=18/5+32/5=10.
I made the mistake of brute force. In the circle equation solve for y, substitute into linear equation, set derivative equal to zero, solve for x. Find y, substitute both values into linear equation. Result = 10.
@@SyberMath After the substitution, we will get something like acos(A)+bsin(A), whose range is [-sqrt(a²+b²), sqrt(a²+b²)] So maximum value is right extremum...
@@SyberMath Perhaps something along the following lines: Given x²+y²=4 In polar coordinates, x=2cosθ y=2sinθ Hence 3x+4y=3(2cosθ)+4(2sinθ)=6cosθ+8sinθ -- (i) Consider a right-triangle with shorter edges 6 and 8 (the coefficients of cosθ and sinθ in (i)). The length of the hypotenuse = √((6*6)+(8*8)) = 10 Let's write 6cosθ+8sinθ = 10[(6/10)cosθ+(8/10)sinθ] -- (ii) Suppose sinβ=6/10; then cosβ=8/10 Substituting these in (ii), we get: 3x+4y=10[sinβcosθ+cosβsinθ]=10sin(β+θ) -- (iii) Thus, the maximum of 3x+4y=maximum of 10sin(β+θ) = 10 (because the maximum value of sin(β+θ) = 1)
You can also actually draw a circle with the centre being the origin and the radius being 2. Then shift a line whose slope is -3/4 to where the line is tangential to the circle. There should be 2 tangential point, the position of the upper one is the x and y of the answer, substituting them yields the maximum.
There is a simpler way of doing this by setting up a Lagrange multiplier, such that L = 3x + 4y - m(4 - x^2 - y^2). Now you partially differentiate L wrt x, y, and m respectively and eliminate m to find the solution.
Note: x^2+y^2=r^2 is the equation of a circle. It is also the pythagorean theorem. And with 3 and 4 we can see a 3,4,5 triangle coming up. 3x+4y=c=10 is the equation of the tangent(slope=-3/4). And 4x-3y=0 is the radius that meets the tangent at right angles.. slope=4/3. So you could solve it using y=4x/3 substituted into the equation for the circle (25/9)x^2 =2^2 => x=6/5 so y=8/5, 3x+4y=50/5=10
A way I tried to visualize this is the following: x^2 + y^2 = 4 is a circle with center at (0, 0) and radius 2. Now 4y + 3x = c needs the highest c such that this line will intersect the circle. Now you know this line has to be tangent to the circle and you have two such points which you can test out yourself (as c increases the lines that are formed are parallel to one another if I'm not mistaken)
Yeah. The geometric way is the quickest, indeed. It takes one second to come to the solution: the function f(x, y) :3x+4y =a has the max if a =10 ( tangent the circle)
We can also solve by geometry by considering x^2+y^2=4 as a circle in coordinate plane with origin as the centre and as we know that we have to find max(3x+4y)=k, so let's consider a straight line 3x+4y=k and from graph we know as a,b>0, where a,b are coefficients of x and y so we find that k is maximum when the line is at the maximum distance from origin and the slope is negative, so to have maximum value of k for a point on the circle, line must much farther that it must meet the circle at one point which means that k must be that the line touches the circle and the we can use distance of line from origin=radius of circle as origin is the centre of the circle and we can get the max value of k
I argued that the dot product of (3,4) and (x,y) reaches its maximum value when (x,y) is a positive multiple of (3,4). Because of the condition, the length must be 2, so I took (2/5)×(3,4) and put the resulting values into the expression 3x+4y
I've never seen something like this before. It's right up my alley for what i am looking for. The whole concept of elementary algebra concepts "growing up" and transforming into higher Mathematics. I'm excited to watch this video!!
One more approach would be to assume a vector A=3i+4j and B=xi+yj A.B=|A| × |B| ×cos∅, where ∅ is the angle between A and B. So (3i+4y)(xi+yj)=√3²+4² × √x²+y² ×cos∅ 3x+4y=5√4cos∅ 3x+4y=10cos∅. For 3x+4y to be maximum, cos ∅ should be maximum and range of cos ∅ is -1≤cos∅≤1. So (cos ∅)max=1 So (3x+4y)max=10
Yes, of course! It's easy to find that the function f(x,y)=3x+4y has a maximum on the curve g(x,y)=x^2+y^2-4=0 at the point (6/5,8/5) the value 10. And by symmetry it has a minimum if you reflect the point on the origin. By the way this channel is awesome! It's fun to see different methods!
What I did was solve x²+y²=4 for y (positive square root), plug it into 3x+4y, differentiate wrt x and set equal to 0. Find x, find y, and compute 3x+4y = 10
I approached it with polar co-ordinates: r^2=4 and we want to maximize f(theta)=3*r*cos(theta) + 4*r*sin(theta). Firstly, r=+(4)^1/2=2. Secondly, f'(theta)=4*r*cos(theta) - 3*r*sin(theta), so the maximum for f should be at the point where 4*r*cos(theta) - 3*r*sin(theta)=0, therefore 8*cos(theta)=6*sin(theta), therefore sin(theta)=(8/6)*cos(theta)=(4/3)*cos(theta), therefore theta=Arctan(4/3), at which point sin(theta)=4/5 and cos(theta)=3/5, therefore y=r*sin(theta)=2*4/5=1.6 and x=r*cos(theta)=2*3/5=1.2 and f(theta)=4*1.6 + 3*1.2=6.4+3.6=10
@@SyberMath Hall and Knight Higher Algebra book, misspelled in my previous comment. I think you have seen this book. You can google it. Well thanks for liking my comment.
Let x = 2*sin a. Then y = 2*cos a. Let z = 3x + 4y = 5*(3/5*2*sin a + 4/5*cos a) = 10*sin (a + b) where cos b = 3/5 & sin b = 4/5. As the max value of sin(a+b) is one, the max value of z = 10*1 = 10. Note that the 5 in the expression for z = square root (3^2 + 4^2) = 5
Hi, alternative method: if x^2+y^2 = 4 means that (x,y)€circle with radius 2. We can then maxmimize : 3*2*cos(t) + 4*2*sin(t) [x=2*cos(t); y=2*sin(t)] and we derivate, we have then : 4*cos(t) = 3*sin(t) By back substitution : 10*cos(t) ; cos(t) our max is 10
I solved it by expressing the 4y term in terms of x, then equating the derivative of the expression to be maximized to 0 and then observing that for x=6/5 the expression indeed reached the maximum of 10. The approach that I used requires the knowledge of how to calculate derivatives, yours with the accompanying reasoning is great.
Using lazy derivretives: y^2=4-x^2 y=sqrt(4-x^2) (3x+4sqrt(4-x^2))’=0 3-8x/2sqrt(4-x^2)=0 -4x/sqrt(4-x^2)=3 16x^2/(4-x^2)=9 t=x^2 16t/(4-t)=9 16t=36-9t 25t=36 t=36/25 x=+-6/5, we of course want it to be positive y^2=4-36/25=(100-36)/25=64/25 y=8/5 3x+4y=18/5+32/5=10
You can also solve it using vector dot-product as follows: 3x+4y can be written as the dot-product of vectors (3i+4j) and (xi+yj) where i and j are unit vectors but dot product of vectors a and b (i.e,, a.b)=|a||b|cosθ where θ is the angle between the vectors Thus, for any two vectors a and b, the dot product is maximum when the angle between them = 0 (because cosθ assumes its maximum 1 when θ=0) Thus, maximum value of a.b = |a||b| Thus, the maximum value of 3x+4y = maximum value of (3i+4j).(xi+yj) = |3i+4j||xi+yj| but |3i+4j|=√(3²+4²)=5 and |xi+yj|=√(x²+y²)=√4=2 (because it's given that x²+y²=4) Thus, the maximum value of 3x+4y = 5 times 2 = 10
Geometry solves it quickly. The locus of the points satisfying x^2+y^2=4 is just a circle of radius 2 centered on the origin. The locus of points where 3x+4y is a constant are lines with slope -3/4. Maximizing this is to simply pick a line that's furthest to the right. Therefore, you are looking for the point in the first quadrant of the circle whose tangent has a slope of -3/4. With geometry you can find that the ray from the origin to this point has tangent 4/3, so it must be (6/5,8/5), and the maximum is 10.
Maximim increase is gradient direction that is 3ux+4uy vector, this direction intersect circle at x=1.2 and y=1.6, substitute these on 3x+4y, you get 10
Since we know that x^2+y^2=4 (1) is a circle with the origin being the center and radius=2, then the sum 3x+4y (2) can maximize only for values of x and y of the first quadrant. So x,y>0. After that we substitute the value of y to (2) and it becomes 3x+4*sqrt(4-x^2) and using calculus we let f(x)=3x+4*sqrt(4-x^2), Df=[0,2]. We find the first derivative to point out possible maxima where the first derivative becomes zero. This happens at x=3/5. We can check if this is a maximum by finding the sign of the first derivative or by finding the second derivative which is negative so f is concave down. We find the maximum straight from calculating the value of f at x=3/5 or we find the value of y from (1) which is y=8/5 and we substitute the two values in the (2). Both of the solutions give us 10.
Besides the algebraic and the trigonometric there is one more visualization of this problem. If the maximum is m, 3x+4y=m then this is obtained when the line y=-3x/4+m/4 with slope -3/4 is tangent of the circle. Differentiating x^2+y^2=R^2 we get xdx+ydy=0, the slope of the tangent to the circle at any point is dy/dx=-x/y=-3/4 and that is it.
There's one more way to solve it, that's using the parametric equation of circle Which leads to an expression of 6cos(a)+8sin(a) which can have a maxima of (6^2+8^2)^0.5
This problem can also be solved using quadratic equations. So let's assume 3x + 4y = k(some constant real number), where x^2 + y^2 = 4, so we can write y = (3 / 4)x - (k / 4). Substituting this in the equation x^2 + y^2 = 4, we finally get 25(x^2) - 6kx + (k^2 - 64) = 0. Since, x is the unknown real number and k is the constant, solution of the equation being real or complex depends upon the discriminant D >= 0, so we have a = 25, b = -6k, and c = k^2 - 64. Therefore, b^2 - 4ac >= 0 means (-6k)^2 - 4 * 25 * (k^2 - 64) >= 0 or -64(k^2) + 6400 >= 0 or k^2
x^2 + y^2 = 4 is a 2-unit circle, so x = 2cosT and y = 2sinT such that x^2+y^2 = 4(cos^2 T + sin^2 T) = 4. Now 3x + 4y = 6cosT + 8sinT = 10 sin(T + arcTan(6/8)). Max of sin or cos = 1. So, Max (3x+4y) = 10 x (max of sin or cos) = 10 x 1 = 10. Hence, max (3x+4y) = *10* . Simple, right ?
A lot of people are talking about lagrange multipliers. Here's another approach that only uses calc 1: turn it into a single variable optimization with trigonometry by parameterizing x=4cos, y=4sin, and then take the derivative of 12cost + 16sint, look for zeros of 16cost-12sint, so tant=4/3, and substitute back in to x=4cost y=4sint using trigonometric identities to get 2 solutions for max and min. Second derivative test then proves which one is max
This is also part of linear algebra or quadratic forms and linear transformation. You are linearly transforming circle into ellipse (both eigenvalues are positive), then the square of maximum has meaning as the sum of squares of eigenvalues or also some of squares of all entries in transformation matrix. That is square of radius of Director circle of the ellipse.
Hi there ! Thanks for your channel. It keeps me in touch with mathematics even though i am far away from it ! We can alse use Lagrangien optimisation. we ´ll then find that maximum are reached for x=6/5 and y=8/5 then 3x+4y will give us 10
Consider the family of parallel straight lines 3x+4y+c=0. x and y lie simultaneously on the line and the circle x²+y²=4, centered at origin. The perpendicular distance of the line from the origin must be ≤ 2 units for the line to intersect, or at least, 'touch' the circle. Mathematically, |c|/5 ≤ 2 or |3x+4y|≤10 or -10≤3x+4y≤10.
@@kartikeyanand9617 to make question more easy.. you just have to imagine how line will move on axis... You will find at point of tangency only max value can be found..
@@kartikeyanand9617 Actually when I saw 3x+4y, which is a linear expression in x and y, I thought of straight lines. Family of straight lines makes things easy. 🙂
Well seen! Cauchy-Schwarz inequality is certainly a tool much more powerful than what is required here. But it makes the whole affair basically a one liner ...
If anyone needs a visual of this: By Cauchy-Schwarz, we have (3^2+4^2)(x^2+y^2) = 25*4 ≥ (3x+4y)^2 → 10 ≥ 3x+4y, and equality is achieved when x/3=y/4, hence the answer is 10.
1. You can express y in terms of x: y= ±sqrt (4-x^2) and investigate the function of one variable f(x)=3*x± 4* sqrt (4 - x^2) for extremes, provided -2≤x≤2. 2. You can use the method of indeterminate Lagrange multipliers - this is the main method for solving such problems. Without any trick
I did the problem using Lagrange multipliers, although solving for y in terms of x, then maximizing by setting the derivative equal to zero would have been just as simple. The maximum value is 10. Actually, try it the second way. y = sqrt(4 - x²) ==> dy/dx = -x/y. 0 = 3 + 4 dy/dx = 3 - 4x/y ==> 3y = 4x. 4 = x² + y² = x² + (4x/3)² = (9x² + 16x²)/9 = 25x²/9. x² = 36/25, x = 6/5. y = 4x/3 = 8/5. 3x + 4y = (18 + 32)/5 = 10
Good evening! I am from Brazil. I notíced tath the purpose of solve this problem is not to use calculus. So I get 3x+4y - C=0 and we want to maximise C with the restriction x^2+y^2=4. So we want the Max C such the line 3x + 4y -C =0 intercepts the circunference x^2+y2=4. This occurs when the line tangents the circunference. So the distance from (0,0) to the line is equal to 2. Then |C|/5=2 and C= + - 10. So C=10 is the maximum.
Much niecr solution can be obtained graphically, if we get 3x+4y = z we can get straight line equation y = -3/4x + z/4 and we know we want to maximize z. So the question is what is the maximum value of z that the line still has intersection with a circle. Or maybe what are extreme cases, when the line has only one intersection point with a circle (at z=-10 or z= 10)
Thanks for the solution. I did this way: we want to find out m such that the circle x²+y²=4 and the line y=(-3x+m)/4 intersect. Consequently, m,x,y>0 and the line is tangent to the circle and intersects it in only one point, thus the discriminant of x^2+((-3x+m)²/16)=4 equals zero. The rest is calculation.
@SyberMath I appreciate that you did it without trig or irrational numbers. Here's how I did it: Parameterize the circle as (x,y) = 2/(1+t^2) * (1 - t^2, 2t). Maximize P = 3x + 4y which equals (3(1-t^2) + 4(2t))(2/(1+t^2)), which simplifies to P(t) = (-6t^2 + 16t + 6)/(1+ t^2). Set P' = 0, and simplify down to 2t^2 + 3t - 2 = 0. That factors as (2t - 1)(t + 2) = 0. So t = 1/2 or t = -2 Plug both values into P and see which is bigger. P(1/2) = 10; P(-2) = -10. So, the max is 10.
It's based on the Pythagorean formula in one variable: (1-t^2)^2 + (2t)^2 = (1+t^2)^2 Divide both sides by the right-hand side. Then match it up with x^2 + y^2 = 1. And then scale (x,y) accordingly with the radius.
Could we do it like this? Let k=3x+4y, solving for y we get y=(k-3x)/4. Substituting for y in the original equation x^2+y^2=4, we get that x^2+(k^2-6kx+9x^2)/16 = 4. Multiplying the whole thing by 16, we get 16x^2+k^2-6kx+9x^2=64 and if we simplify we get 25x^2-6kx+k^2-64=0. Assuming x is real, we need the discriminant to be nonnegative so we have that 36k^2-4(25)(k^2-64)>=0 or -64k^2+6400>=0. We factor to get -64(k^2-100) >=0 and divide by -64 to get k^2-100
From x^2 + y^2 = 4, first we can get y from it, which is sqrt(x^2 - 4), substitute it into 3x+4y, and get 3x + 4(sqrt(x^2 - 4)), get the derivative of that and get the critical points by setting it to zero. And when we're done we get x= 6/5 or x = -6/5 We do the same for y, aka we get x from the first equation, and then substitute it into the second one, and we get: y = 8/5 or y = -8/5 Now since we're adding and multiplying by positive numbers, the higher X and Y are, the higher the final value is. Therefore the maximum is X = 6/5 and Y = 8/5 And when we substitute those values into the equation we get 10 Solved using calculus
Alternative quick solution: Let u and v be real vectors. Then, max(u•v) = |u|*|v|. This is simply true from the definition of the dot product: a•b = |a|*|b|*cos(θ). Clearly, for the dot product to be maximized, cos(θ) must be 1 (i.e., the two vectors must be parallel and are facing the same direction). That fact can be used for the problem by letting u be a vector that represents a point on a circle (which in this case, has a radius of 2), and let v be the vector [3,4]. So, we see that max(u•v) = 2*5 = 10. Problem solved!
the maximum value occurs at 4y=3x in which case x^2 +y^2 =4 need not necessarily hold good. Its value can not cross 100 it doesnt mean it can touch 100.
Using the graphical solutions for this I got that the maximum value of x is 2 and the maximum value of Y is also 2 so I plugged that in the expression 3 X + 4 Y to get its maximum value and I got 14 so what do you think about that answer.
I intuit that multiplying circle coordinates by coefficients will be a maximum when each coordinate result is equivalent. Thus if 3x=4y, it quickly results in the values for x, y (6/5,8/5) and the result is 10. In essence, you proved this fact in the algebraic property of the sum of squares. Found this interesting to think about.
Ok, that was very impressive. However there's one thing that I don't get: I know (4x - 3y)² in the end can't be negative, but isn't it necessary to prove its lowest value is, in fact, 0? Could someone help me?
You are correct. It's not very hard to show: Smallest value of (4x-3y)^2 is 0. From 4x-3y=0 follows y=4x/3. Substitute this into x^2 + y^2 = 4 and you get x=6/5 and also y=8/5. Negative solutions for x, y can be disregared because they do not provide the maximum.
Well, the answer is yes AND no, technically you should prove it. But it's clear that it would be zero when x and y are in the ratio of 3:4, and we can see that there must exist some values which satisfy that AND x^2+y^2=4 as it is essentially just the equation of a circle
The ratio of x to y for the answer is 3:4 so is that a coincidence or is it related to 3x+4y that we are trying to maximize? If it is related, then we have more information. I wrote a computer program checking all values for x and y starting at 0 and incrementing by 0.01, but if I knew the answer to x and y was in a 3:4 ratio, I could instead make y always equal to 4/3rd that of x.
Interesting formula and, as always, elegant solution. I prefered something more obvious: f(x)=3x+4sqrt(4-x^2), f'(x)=3+4*(-2x)/2sqrt(4-x^2)=0, 3-4x/sqrt(4-x^2)=0, 3=4x/sqrt(4-x^2), 4x=3sqrt(4-x^2), 16x^2=36-9x^2, x^2=36/25, x=+/-6/5. y=sqrt(4-36/25)=sqrt(64/25)=+/-8/5. Taking both postive values we are left with 3*6/5+4*8/5=18/5+32/5=50/5=10. Technically, it's a little illegal, because using the square root, I use only a half of the domain for x and y. But in this case, positive values give us a bigger value of 3x+4y, so I got away with it.
Very nice example, it has geometrical meaning of defining radius of the Director circle of ellipse defined by semiaxis a and b and also it represents energy of two dimensional oscillator or coupled oscillators of the same frequency. It is also representing amplitude of orthogonally combined cosine and sine of the same frequency/speed factor in Fourier series.
@@anshumanagrawal346 Since we know that x^2+y^2=4 (1) is a circle with the origin being the center and radius=2, then the sum 3x+4y (2) can maximize only for values of x and y of the first quadrant. So x,y>0. After that we substitute the value of y to (2) and it becomes 3x+4*sqrt(4-x^2) and using calculus we let f(x)=3x+4*sqrt(4-x^2), Df=[0,2]. We find the first derivative to point out possible maxima where the first derivative becomes zero. This happens at x=3/5. We can check if this is a maximum by finding the sign of the first derivative or by finding the second derivative which is negative so f is concave down. We find the maximum straight from calculating the value of f at x=3/5 or we find the value of y from (1) which is y=8/5 and we substitute the two values in the (2). Both of the solutions give us 10.
@@geosalatast5715 Oh, I thought of solving this with maxima/minima as well but thought it was too long. When the original comment said they used calculus I thought it was some sort of some directly find the maximum of 3x+4y using partial derivatives
A more geometric approach: Let's consider a point M(x,y). Obviously, the point M lies on the circle with center at the origin and radius 2. We want to find the maximum value of 3x+4y. Let 3x+4y=a. This equation represents a straight line in the xy plane. The point M lies on the line, too. So we want the circle and line to have common points. That happens if-f the distance of the center from the line is less than or equal to the radius. From the formula of the distance between a point and a line we get |a|≤10. So the maximum value is 10. It is achieved when the line (for a=10) is tangent to the circle, that is at the point (6/5,8/5).
I solved this with analysis - get x from that circle without +/- because during f'(x) = 0 you will be turning it into square anyway and get the same results. You will get y = +/- 8/5 . I'm trying to find algebraic or geometric solutions, but when i know easier way my brain stops working.
x=2cost, y=2sin(t) 6cos(t)+8sin(t)=10(0.6cos(t)+0.8sin(t)) =10cos(t - theta) cos(theta) = 0.6 sin(theta) = 0.8 max = 10 for angle t=theta+2kπ or t=-theta+2kπ where k in Z
Very elegant. But why run through the specific calculation after you have just proved the general case? Just substitute for a and b in the general formula. Are you trying to make the video longer? ;)
I appreciate this algebric trick but here is a derivation using trigonometric identites. As you mention in another response to a comment it boils down to maximizing (6costheta + 8sintheta), but this also reduces to maximizing cos(theta)+tan(phi)*sin(theta) where phi is the angle with tangent 4/3. Then this can be factored to read max(sin(phi+theta)) which is maximum for phi+theta=pi/2 and this means that tan(theta)=cot(phi)=3/4=x/y as in the condition you obtained algebrically and in some way justifies the inversion of the coefficients.
My solution (of course extreeeeeeeeeemely faster): Sometimes I use algebra to solve a geometrical problem, this time I do the opposite. x²+y²=4 is the equation of the circle centered on O of coordinates (0,0) and of radius 2. 3x+4y=K, where K is a real number, is the equation of a line. We're looking at the maximum value of K where the line intersects the circle. We can see that when K rises, our lines moves upwards. If K is too low or too high, there are no points of the line that are also on the circle and when K rises inside the range where the line and the circle meet each other, the common points move to the north-east. So the value of K that we're looking for is where the line is tangent to the circle in the north-east. But since the equation of the line is 3x+4y=K and since the radius is coming from O, we know that the radius belongs to the line of equation: y=4/3.x. We report this into the equation of the circle: 25/9.x²=4 => x²=36/25 => x=6/5 (x is positive). y=4/3.(6/5)=8/5. Then 3x+4y=18/5+32/5=10.
Divide the equation by 4 both sides, then write it as (x/2)^2+(y/2)^2=1 then let x/2=sin theta so y/2=cos theta. Sub x=2sin theta and y=2 cos theta in the 2nd equation we get max(6sin theta +8cos theta). And, max and min value of a sin theta+ b cos theta is ± √(a^2+b^2) So √(6^2+8^2)=√100=10
Most of the time you are taking a very lengthy approach to solve this kind of easy problems. I think taking an easy approach will make the students interesting. Consider the family of parallel lines 3x + 4y = c; the idea is to maximize the value c. there will be two parallels in this family tangential to the circle x2 + y2 = 4 and giving maximum value of c which will be 10.
I find your videos great, but it would be very helpful for me, if you tried to explain your plan before giving „random“ identities. I did not really get the sense of the identity before the end. Thanks for helping me to get a better intuition on algebraic problem-solving!
Easy trig. substitution or by method of vectors, this q. can be solved. But your approach is formal & leaves no scope for any domain kind of a mistake! Congrats!!!
Let's solve this problem using Lagrange multipliers. Let F(x,y) = 3x + 4y, subject to the constraint G(x,y) = x^2 + y^2 - 4 = 0. Fx = λGx & Fy = λGy, by theorem finds the extreme points, where λ is real and Fx is the partial derivative of F wrt to x, Fy is the partial derivative of F wrt to y, Gx is the partial derivative of G wrt to x, Gy is the partial derivative of G wrt to y. Now Fx = 3 & Fy = 4 and Gx = 2x & Gy = 2y. So, 3 = λ(2x) & 4 = λ(2y) So, 9 = 4(λ^2)(x^2) & 16 = 4(λ^2)(y^2) So, 25 = 4(λ^2)[x^2 + y^2] So, 25 = 4(λ^2)[4], as x^2 + y^2 = 4 So, 25 = 16(λ^2) So, λ = ±5/4 leading to (x,y) = (±6/5, ±8/5) as the extreme points, using 3 = λ(2x) & 4 = λ(2y), which results in the maximum of 3x + 4y being 10 when (x,y) = (6/5, 8/5), and the minimum of 3x + 4y being -10 when (x,y) = (-6/5, -8/5)
Hii mst syber !! As always amazing problems , until now i didn t watch your approach , i think my approach is more complicated than your s I let y=k.x then let 3x + 4y = t Now we must maximize t if you plug the substitution in the equations you get a system ..... finaly you get a function of k t(k)=...... You take the derivitive and you get one solution which is max ......... finaly you get t or (3x+4y) =10
![BLACK BAG - Official Trailer [HD] - Only in Theaters March 14](http://i.ytimg.com/vi/Du0Xp8WX_7I/mqdefault.jpg)
Here is Riley's video on Lambert-W function! Check it out:
ruclips.net/video/Go5W4Nacn3Q/видео.html
I can solve this problem when I put x=2sina, y=2cosa then equation is satisfied. And I put value of x and y in last equation like 6sina+8cosa then Max value is √6^2+8^2=10 . I wait for you your reply. I am a jee student from India🇮🇳
you are teaching maths? in scholl mr sybermath?
so you take maximum 3x+4y = (3² + 4²) (x²+y²) = 100; then you have system of 2 equasions and can get x and y; for example x = 6/5
Since x, y lies on the circle with radius 2 we can write x=2cost,y=2sint now 3x +4y becomes 6cost+8sint and we know that max. value of acosx+bsinx is √(a^2+b^2) so the max. value of our expression will be ✓(6^2+8^2)=10
I thought so too.
As soon as I saw the equation, I thought that the traces of x,y were circles.
Definitely more refined than either of my methods.
I used a similar method but I didn't use trigonometry, just equations of lines:
x²+y²=4 is the equation of the circle centered on O of coordinates (0,0) and of radius 2.
3x+4y=K, where K is a real number, is the equation of a line. We're looking at the maximum value of K where the line intersects the circle.
We can see that when K rises, our lines moves upwards. If K is too low or too high, there are no points of the line that are also on the circle and when K rises inside the range where the line and the circle meet each other, the common points move to the north-east.
So the value of K that we're looking for is where the line is tangent to the circle in the north-east. But since the equation of the line is 3x+4y=K and since the radius is coming from O, we know that the radius belongs to the line of equation: y=4/3.x.
We report this into the equation of the circle: 25/9.x²=4 => x²=36/25 => x=6/5 (x is positive).
y=4/3.(6/5)=8/5.
Then 3x+4y=18/5+32/5=10.
I made the mistake of brute force. In the circle equation solve for y, substitute into linear equation, set derivative equal to zero, solve for x. Find y, substitute both values into linear equation. Result = 10.
Using Cauchy inequality:（x^2+y^2)(9+16)≥(3x+4y)^2 , (3x+4y)^2≤100 ,3x+4y≤10 ,so that the maximum is 10
Using trigonometric substitution, we can directly see the answer to be 10.
How?
@@SyberMath substituting x=2costheta and y=2sintheta leaving us with max(6costheta + 8sintheta) and the sqrt(6^2+8^2) = 10
Same
@@SyberMath After the substitution, we will get something like acos(A)+bsin(A), whose range is [-sqrt(a²+b²), sqrt(a²+b²)]
So maximum value is right extremum...
@@SyberMath Perhaps something along the following lines:
Given x²+y²=4
In polar coordinates,
x=2cosθ
y=2sinθ
Hence 3x+4y=3(2cosθ)+4(2sinθ)=6cosθ+8sinθ -- (i)
Consider a right-triangle with shorter edges 6 and 8 (the coefficients of cosθ and sinθ in (i)). The length of the hypotenuse = √((6*6)+(8*8)) = 10
Let's write 6cosθ+8sinθ = 10[(6/10)cosθ+(8/10)sinθ] -- (ii)
Suppose sinβ=6/10; then cosβ=8/10
Substituting these in (ii), we get:
3x+4y=10[sinβcosθ+cosβsinθ]=10sin(β+θ) -- (iii)
Thus, the maximum of 3x+4y=maximum of 10sin(β+θ) = 10 (because the maximum value of sin(β+θ) = 1)
You can also actually draw a circle with the centre being the origin and the radius being 2. Then shift a line whose slope is -3/4 to where the line is tangential to the circle. There should be 2 tangential point, the position of the upper one is the x and y of the answer, substituting them yields the maximum.
There is a simpler way of doing this by setting up a Lagrange multiplier, such that L = 3x + 4y - m(4 - x^2 - y^2). Now you partially differentiate L wrt x, y, and m respectively and eliminate m to find the solution.
Note: x^2+y^2=r^2 is the equation of a circle. It is also the pythagorean theorem. And with 3 and 4 we can see a 3,4,5 triangle coming up.
3x+4y=c=10 is the equation of the tangent(slope=-3/4). And 4x-3y=0 is the radius that meets the tangent at right angles.. slope=4/3.
So you could solve it using y=4x/3 substituted into the equation for the circle (25/9)x^2 =2^2 => x=6/5 so y=8/5, 3x+4y=50/5=10
Nice!
A way I tried to visualize this is the following:
x^2 + y^2 = 4 is a circle with center at (0, 0) and radius 2. Now 4y + 3x = c needs the highest c such that this line will intersect the circle. Now you know this line has to be tangent to the circle and you have two such points which you can test out yourself (as c increases the lines that are formed are parallel to one another if I'm not mistaken)
Yeah.
The geometric way is the quickest, indeed.
It takes one second to come to the solution: the function f(x, y) :3x+4y =a has the max if a =10 ( tangent the circle)
We can also solve by geometry by considering x^2+y^2=4 as a circle in coordinate plane with origin as the centre and as we know that we have to find max(3x+4y)=k, so let's consider a straight line 3x+4y=k and from graph we know as a,b>0, where a,b are coefficients of x and y so we find that k is maximum when the line is at the maximum distance from origin and the slope is negative, so to have maximum value of k for a point on the circle, line must much farther that it must meet the circle at one point which means that k must be that the line touches the circle and the we can use distance of line from origin=radius of circle as origin is the centre of the circle and we can get the max value of k
That's cool!
genial
I argued that the dot product of (3,4) and (x,y) reaches its maximum value when (x,y) is a positive multiple of (3,4). Because of the condition, the length must be 2, so I took (2/5)×(3,4) and put the resulting values into the expression 3x+4y
I've never seen something like this before. It's right up my alley for what i am looking for. The whole concept of elementary algebra concepts "growing up" and transforming into higher Mathematics. I'm excited to watch this video!!
Very glad to hear that! ☺️
For proof that 10 is attainable, use x=6/5 and y=8/5.
That's good!
Adding the important bit. :)
optimization problem, nice:
*Maximize[{3 x + 4 y, x^2 + y^2 == 4}, {x, y}]*
Hello computer! 🤗😂
One more approach would be to assume a vector A=3i+4j and B=xi+yj
A.B=|A| × |B| ×cos∅, where ∅ is the angle between A and B.
So
(3i+4y)(xi+yj)=√3²+4² × √x²+y² ×cos∅
3x+4y=5√4cos∅
3x+4y=10cos∅. For 3x+4y to be maximum, cos ∅ should be maximum and range of cos ∅ is -1≤cos∅≤1. So (cos ∅)max=1
So
(3x+4y)max=10
We can also use lagrangian multipliers, right? Or am I wrong?
Indeed, and you in fact, this was my first thought as well. Having said that, it's interesting to see a different approach.
You can!
Right
Yes, of course! It's easy to find that the function f(x,y)=3x+4y has a maximum on the curve g(x,y)=x^2+y^2-4=0 at the point (6/5,8/5) the value 10. And by symmetry it has a minimum if you reflect the point on the origin. By the way this channel is awesome! It's fun to see different methods!
x=r*cos alfa,y=r*sin alfa;there is maximum for r=2; alfa=arctg(4/3)
What I did was solve x²+y²=4 for y (positive square root), plug it into 3x+4y, differentiate wrt x and set equal to 0. Find x, find y, and compute 3x+4y = 10
Nice!
Yeah, i did the same.
I approached it with polar co-ordinates: r^2=4 and we want to maximize f(theta)=3*r*cos(theta) + 4*r*sin(theta). Firstly, r=+(4)^1/2=2.
Secondly, f'(theta)=4*r*cos(theta) - 3*r*sin(theta), so the maximum for f should be at the point where 4*r*cos(theta) - 3*r*sin(theta)=0, therefore
8*cos(theta)=6*sin(theta), therefore
sin(theta)=(8/6)*cos(theta)=(4/3)*cos(theta), therefore
theta=Arctan(4/3), at which point sin(theta)=4/5 and cos(theta)=3/5, therefore
y=r*sin(theta)=2*4/5=1.6 and x=r*cos(theta)=2*3/5=1.2 and f(theta)=4*1.6 + 3*1.2=6.4+3.6=10
Most of your questions are from H.S haul book, I like this book. The questions are very hard in this. Your solutions are awesome.
Which book is that?
@@SyberMath Hall and Knight Higher Algebra book, misspelled in my previous comment. I think you have seen this book. You can google it. Well thanks for liking my comment.
@@SyberMath I think he meant the famous Text Book called School Algebra by H. S Hall and Night published end of 1890s and many reprints till date.
Let x = 2*sin a. Then y = 2*cos a.
Let z = 3x + 4y = 5*(3/5*2*sin a + 4/5*cos a) = 10*sin (a + b) where
cos b = 3/5 & sin b = 4/5. As the max value of sin(a+b) is one, the max value of z = 10*1 = 10.
Note that the 5 in the expression for z = square root (3^2 + 4^2) = 5
Wow! That's very cool. Focused so much on this identity, I did not even think about this 😁
Your solutions are wonderful. I am a big fan of your videos! Keep posting such wonderful problems & solutions. Thanks.
De todas las respuestas está me encantó
Hi, alternative method: if x^2+y^2 = 4 means that (x,y)€circle with radius 2.
We can then maxmimize : 3*2*cos(t) + 4*2*sin(t) [x=2*cos(t); y=2*sin(t)]
and we derivate, we have then : 4*cos(t) = 3*sin(t)
By back substitution : 10*cos(t) ; cos(t) our max is 10
Glad I found this channel this is so good.
Glad you enjoy it!
I solved it by expressing the 4y term in terms of x, then equating the derivative of the expression to be maximized to 0 and then observing that for x=6/5 the expression indeed reached the maximum of 10.
The approach that I used requires the knowledge of how to calculate derivatives, yours with the accompanying reasoning is great.
Using lazy derivretives:
y^2=4-x^2
y=sqrt(4-x^2)
(3x+4sqrt(4-x^2))’=0
3-8x/2sqrt(4-x^2)=0
-4x/sqrt(4-x^2)=3
16x^2/(4-x^2)=9
t=x^2
16t/(4-t)=9
16t=36-9t
25t=36
t=36/25
x=+-6/5, we of course want it to be positive
y^2=4-36/25=(100-36)/25=64/25
y=8/5
3x+4y=18/5+32/5=10
You can also solve it using vector dot-product as follows:
3x+4y can be written as the dot-product of vectors (3i+4j) and (xi+yj) where i and j are unit vectors
but dot product of vectors a and b (i.e,, a.b)=|a||b|cosθ where θ is the angle between the vectors
Thus, for any two vectors a and b, the dot product is maximum when the angle between them = 0 (because cosθ assumes its maximum 1 when θ=0)
Thus, maximum value of a.b = |a||b|
Thus, the maximum value of 3x+4y = maximum value of (3i+4j).(xi+yj) = |3i+4j||xi+yj|
but |3i+4j|=√(3²+4²)=5
and |xi+yj|=√(x²+y²)=√4=2 (because it's given that x²+y²=4)
Thus, the maximum value of 3x+4y = 5 times 2 = 10
Geometry solves it quickly. The locus of the points satisfying x^2+y^2=4 is just a circle of radius 2 centered on the origin. The locus of points where 3x+4y is a constant are lines with slope -3/4. Maximizing this is to simply pick a line that's furthest to the right.
Therefore, you are looking for the point in the first quadrant of the circle whose tangent has a slope of -3/4. With geometry you can find that the ray from the origin to this point has tangent 4/3, so it must be (6/5,8/5), and the maximum is 10.
Maximim increase is gradient direction that is 3ux+4uy vector, this direction intersect circle at x=1.2 and y=1.6, substitute these on 3x+4y, you get 10
Since we know that x^2+y^2=4 (1) is a circle with the origin being the center and radius=2, then the sum 3x+4y (2) can maximize only for values of x and y of the first quadrant. So x,y>0. After that we substitute the value of y to (2) and it becomes 3x+4*sqrt(4-x^2) and using calculus we let f(x)=3x+4*sqrt(4-x^2), Df=[0,2]. We find the first derivative to point out possible maxima where the first derivative becomes zero. This happens at x=3/5. We can check if this is a maximum by finding the sign of the first derivative or by finding the second derivative which is negative so f is concave down. We find the maximum straight from calculating the value of f at x=3/5 or we find the value of y from (1) which is y=8/5 and we substitute the two values in the (2). Both of the solutions give us 10.
Besides the algebraic and the trigonometric there is one more visualization of this problem. If the maximum is m, 3x+4y=m then this is obtained when the line y=-3x/4+m/4 with slope -3/4 is tangent of the circle. Differentiating x^2+y^2=R^2 we get xdx+ydy=0, the slope of the tangent to the circle at any point is dy/dx=-x/y=-3/4 and that is it.
There's one more way to solve it, that's using the parametric equation of circle Which leads to an expression of 6cos(a)+8sin(a) which can have a maxima of (6^2+8^2)^0.5
This problem can also be solved using quadratic equations. So let's assume 3x + 4y = k(some constant real number), where x^2 + y^2 = 4, so we can write y = (3 / 4)x - (k / 4). Substituting this in the equation x^2 + y^2 = 4, we finally get 25(x^2) - 6kx + (k^2 - 64) = 0. Since, x is the unknown real number and k is the constant, solution of the equation being real or complex depends upon the discriminant D >= 0, so we have a = 25, b = -6k, and c = k^2 - 64. Therefore, b^2 - 4ac >= 0 means (-6k)^2 - 4 * 25 * (k^2 - 64) >= 0 or -64(k^2) + 6400 >= 0 or k^2
x^2 + y^2 = 4 is a 2-unit circle, so x = 2cosT and y = 2sinT such that x^2+y^2 = 4(cos^2 T + sin^2 T) = 4. Now 3x + 4y = 6cosT + 8sinT = 10 sin(T + arcTan(6/8)). Max of sin or cos = 1. So, Max (3x+4y) = 10 x (max of sin or cos) = 10 x 1 = 10. Hence, max (3x+4y) = *10* . Simple, right ?
Your method of solving a problem is amazing. It requires a lot of knowledge and reasoning. Thank you.
Glad you think so!
A lot of people are talking about lagrange multipliers. Here's another approach that only uses calc 1: turn it into a single variable optimization with trigonometry by parameterizing x=4cos, y=4sin, and then take the derivative of 12cost + 16sint, look for zeros of 16cost-12sint, so tant=4/3, and substitute back in to x=4cost y=4sint using trigonometric identities to get 2 solutions for max and min. Second derivative test then proves which one is max
This is also part of linear algebra or quadratic forms and linear transformation. You are linearly transforming circle into ellipse (both eigenvalues are positive), then the square of maximum has meaning as the sum of squares of eigenvalues or also some of squares of all entries in transformation matrix. That is square of radius of Director circle of the ellipse.
Interesting!
Hi there ! Thanks for your channel. It keeps me in touch with mathematics even though i am far away from it ! We can alse use Lagrangien optimisation. we ´ll then find that maximum are reached for x=6/5 and y=8/5 then 3x+4y will give us 10
Glad to hear that!
Consider the family of parallel straight lines 3x+4y+c=0. x and y lie simultaneously on the line and the circle x²+y²=4, centered at origin. The perpendicular distance of the line from the origin must be ≤ 2 units for the line to intersect, or at least, 'touch' the circle. Mathematically, |c|/5 ≤ 2 or |3x+4y|≤10 or -10≤3x+4y≤10.
why did you considered family of straight lines
@@kartikeyanand9617 to make question more easy.. you just have to imagine how line will move on axis... You will find at point of tangency only max value can be found..
Good thinking! 😍
@@SyberMath Thanks!!!!!
@@kartikeyanand9617 Actually when I saw 3x+4y, which is a linear expression in x and y, I thought of straight lines. Family of straight lines makes things easy. 🙂
A quick application of the caughy Schwartz inequality would have done it but your way is more elegant :)
Well seen! Cauchy-Schwarz inequality is certainly a tool much more powerful than what is required here. But it makes the whole affair basically a one liner ...
If anyone needs a visual of this:
By Cauchy-Schwarz, we have
(3^2+4^2)(x^2+y^2) = 25*4 ≥ (3x+4y)^2
→ 10 ≥ 3x+4y, and equality is achieved when x/3=y/4, hence the answer is 10.
You present an interesting solution. I used the Lagrange multiplier method that's introduced in a third semester Calculus course.
I used Lagrange too. Obtained the same result.
1. You can express y in terms of x: y= ±sqrt (4-x^2) and investigate the function of one variable f(x)=3*x± 4* sqrt (4 - x^2) for extremes, provided -2≤x≤2.
2. You can use the method of indeterminate Lagrange multipliers - this is the main method for solving such problems. Without any trick
Good!
I did the problem using Lagrange multipliers, although solving for y in terms of x, then maximizing by setting the derivative equal to zero would have been just as simple. The maximum value is 10.
Actually, try it the second way. y = sqrt(4 - x²) ==> dy/dx = -x/y. 0 = 3 + 4 dy/dx = 3 - 4x/y ==> 3y = 4x.
4 = x² + y² = x² + (4x/3)² = (9x² + 16x²)/9 = 25x²/9. x² = 36/25, x = 6/5. y = 4x/3 = 8/5.
3x + 4y = (18 + 32)/5 = 10
Good evening! I am from Brazil. I notíced tath the purpose of solve this problem is not to use calculus. So I get 3x+4y - C=0 and we want to maximise C with the restriction x^2+y^2=4. So we want the Max C such the line 3x + 4y -C =0 intercepts the circunference x^2+y2=4. This occurs when the line tangents the circunference. So the distance from (0,0) to the line is equal to 2. Then |C|/5=2 and C= + - 10. So C=10 is the maximum.
Good to see you! Greetings from the United States 💖
@@SyberMath, I liked the way you solve.
Much niecr solution can be obtained graphically, if we get 3x+4y = z we can get straight line equation y = -3/4x + z/4 and we know we want to maximize z. So the question is what is the maximum value of z that the line still has intersection with a circle. Or maybe what are extreme cases, when the line has only one intersection point with a circle (at z=-10 or z= 10)
Thanks for the solution. I did this way: we want to find out m such that the circle x²+y²=4 and the line y=(-3x+m)/4 intersect. Consequently, m,x,y>0 and the line is tangent to the circle and intersects it in only one point, thus the discriminant of x^2+((-3x+m)²/16)=4 equals zero. The rest is calculation.
@SyberMath I appreciate that you did it without trig or irrational numbers. Here's how I did it:
Parameterize the circle as (x,y) = 2/(1+t^2) * (1 - t^2, 2t).
Maximize P = 3x + 4y which equals (3(1-t^2) + 4(2t))(2/(1+t^2)), which simplifies to P(t) = (-6t^2 + 16t + 6)/(1+ t^2).
Set P' = 0, and simplify down to 2t^2 + 3t - 2 = 0. That factors as (2t - 1)(t + 2) = 0.
So t = 1/2 or t = -2
Plug both values into P and see which is bigger.
P(1/2) = 10; P(-2) = -10. So, the max is 10.
I don't get how you parametrize the circle 🤔
It's based on the Pythagorean formula in one variable:
(1-t^2)^2 + (2t)^2 = (1+t^2)^2
Divide both sides by the right-hand side.
Then match it up with x^2 + y^2 = 1. And then scale (x,y) accordingly with the radius.
@@SyberMath ruclips.net/video/xp0H3Aw0j6E/видео.html Wildberger explains it very well
@@SyberMath he is using half angle substitution t=θ/2 and express sinθ, cosθ in terms of t (x,y)= (2cosθ, 2sinθ) in the equation of the circle
Consider two vectors a= 3i+4j and b= xi+yj. Consider dot product and maximize when cost =1
Let x=2sinA and y=2cosA then eqn would be 6sinA+8cosA so largest is 10 and lowest -10
What an interesting problem! You are a genius at finding great problems
I used x=2costheta and y=2sintheta to solve this one!!
Thank you! 😍
Could we do it like this? Let k=3x+4y, solving for y we get y=(k-3x)/4. Substituting for y in the original equation x^2+y^2=4, we get that x^2+(k^2-6kx+9x^2)/16 = 4. Multiplying the whole thing by 16, we get 16x^2+k^2-6kx+9x^2=64 and if we simplify we get 25x^2-6kx+k^2-64=0. Assuming x is real, we need the discriminant to be nonnegative so we have that 36k^2-4(25)(k^2-64)>=0 or -64k^2+6400>=0. We factor to get -64(k^2-100) >=0 and divide by -64 to get k^2-100
From x^2 + y^2 = 4, first we can get y from it, which is sqrt(x^2 - 4), substitute it into 3x+4y, and get 3x + 4(sqrt(x^2 - 4)), get the derivative of that and get the critical points by setting it to zero.
And when we're done we get x= 6/5 or x = -6/5
We do the same for y, aka we get x from the first equation, and then substitute it into the second one, and we get:
y = 8/5 or y = -8/5
Now since we're adding and multiplying by positive numbers, the higher X and Y are, the higher the final value is. Therefore the maximum is X = 6/5 and Y = 8/5
And when we substitute those values into the equation we get 10
Solved using calculus
Nice!
Alternative quick solution:
Let u and v be real vectors. Then,
max(u•v) = |u|*|v|.
This is simply true from the definition of the dot product: a•b = |a|*|b|*cos(θ). Clearly, for the dot product to be maximized, cos(θ) must be 1 (i.e., the two vectors must be parallel and are facing the same direction).
That fact can be used for the problem by letting u be a vector that represents a point on a circle (which in this case, has a radius of 2), and let v be the vector [3,4]. So, we see that
max(u•v) = 2*5 = 10.
Problem solved!
the maximum value occurs at 4y=3x in which case x^2 +y^2 =4 need not necessarily hold good.
Its value can not cross 100 it doesnt mean it can touch 100.
Using the graphical solutions for this I got that the maximum value of x is 2 and the maximum value of Y is also 2 so I plugged that in the expression
3 X + 4 Y to get its maximum value and I got 14 so what do you think about that answer.
X and Y can’t both be 2, remember x^2 + y^2 = 4. 2^2 + 2^2 = 8 =/= 4.
Simply awesome. Thanks for the fun and simplicity you brought to the steps.
Np. Thank you for the kind words! 💖
I intuit that multiplying circle coordinates by coefficients will be a maximum when each coordinate result is equivalent. Thus if 3x=4y, it quickly results in the values for x, y (6/5,8/5) and the result is 10. In essence, you proved this fact in the algebraic property of the sum of squares. Found this interesting to think about.
Ok, that was very impressive. However there's one thing that I don't get: I know (4x - 3y)² in the end can't be negative, but isn't it necessary to prove its lowest value is, in fact, 0?
Could someone help me?
no you don't have to prove that r^2 >= 0, given that r is a real number, it's a fact
You are correct. It's not very hard to show: Smallest value of (4x-3y)^2 is 0. From 4x-3y=0 follows y=4x/3. Substitute this into x^2 + y^2 = 4 and you get x=6/5 and also y=8/5. Negative solutions for x, y can be disregared because they do not provide the maximum.
@@MathElite thank you, but I was actually talking about this particular case (4x - 3y)²
@@falknfurter got it. Thank you very much
Well, the answer is yes AND no, technically you should prove it. But it's clear that it would be zero when x and y are in the ratio of 3:4, and we can see that there must exist some values which satisfy that AND x^2+y^2=4 as it is essentially just the equation of a circle
Alternately we might have directly used Cauchy's inequality. abs val (a.b)
Excellent
I used CBS inequality (ax+by)^2
The ratio of x to y for the answer is 3:4 so is that a coincidence or is it related to 3x+4y that we are trying to maximize? If it is related, then we have more information. I wrote a computer program checking all values for x and y starting at 0 and incrementing by 0.01, but if I knew the answer to x and y was in a 3:4 ratio, I could instead make y always equal to 4/3rd that of x.
Using vectors
a(v) =3i +4j, mag(a) =5
b(v) =xi+yj, mag(b) =√x^2+y^2=2
a(v) .b(v)
Suppose (x,y) be a point on a circle of radius 2 put x=2cos@ and y=2sin@ and that will bring you to the end .
Set sinu =3/5, cosu = 4/5, cos v = x/2, sinv = y/2 → max(3x/10 + 4y/10) = max(sinucosv + cosusinv) = max(sin(u + v)) = 1 → max(3x + 4y) = 10 max(sin(u + v)) = 10.
I used the cauchy-schwarz inequality to get (a1b1+a2b2)^2
Definitely the easiest way
Interesting formula and, as always, elegant solution. I prefered something more obvious: f(x)=3x+4sqrt(4-x^2), f'(x)=3+4*(-2x)/2sqrt(4-x^2)=0, 3-4x/sqrt(4-x^2)=0, 3=4x/sqrt(4-x^2), 4x=3sqrt(4-x^2), 16x^2=36-9x^2, x^2=36/25, x=+/-6/5. y=sqrt(4-36/25)=sqrt(64/25)=+/-8/5. Taking both postive values we are left with 3*6/5+4*8/5=18/5+32/5=50/5=10. Technically, it's a little illegal, because using the square root, I use only a half of the domain for x and y. But in this case, positive values give us a bigger value of 3x+4y, so I got away with it.
Good method! Good thinking! 💖
It will be maximum when the line will be tangent to the circle. Hence, 10 is the answer obviously.
Very nice example, it has geometrical meaning of defining radius of the Director circle of ellipse defined by semiaxis a and b and also it represents energy of two dimensional oscillator or coupled oscillators of the same frequency. It is also representing amplitude of orthogonally combined cosine and sine of the same frequency/speed factor in Fourier series.
Interesting
I SWEAR THIS IS SO UNDERRATED
This was the first problem presented by you on this channel that I was able to solve! I used calculus. Thank you very much for your content!
Glad it helped!
I used the fact the given condition is the equation of a circle and used some analytical geometry, but I'm curious how you used calculus to solve this
@@anshumanagrawal346 Since we know that x^2+y^2=4 (1) is a circle with the origin being the center and radius=2, then the sum 3x+4y (2) can maximize only for values of x and y of the first quadrant. So x,y>0. After that we substitute the value of y to (2) and it becomes 3x+4*sqrt(4-x^2) and using calculus we let f(x)=3x+4*sqrt(4-x^2), Df=[0,2]. We find the first derivative to point out possible maxima where the first derivative becomes zero. This happens at x=3/5. We can check if this is a maximum by finding the sign of the first derivative or by finding the second derivative which is negative so f is concave down. We find the maximum straight from calculating the value of f at x=3/5 or we find the value of y from (1) which is y=8/5 and we substitute the two values in the (2). Both of the solutions give us 10.
@@geosalatast5715 Oh, I thought of solving this with maxima/minima as well but thought it was too long. When the original comment said they used calculus I thought it was some sort of some directly find the maximum of 3x+4y using partial derivatives
Bravo!
This skill is just beautiful and fantastic!
You relight my hope of number theory!!!!!
Glad to hear that!
Please, could you solve this problem using the derivatives ? Thank you.
A more geometric approach:
Let's consider a point M(x,y). Obviously, the point M lies on the circle with center at the origin and radius 2. We want to find the maximum value of 3x+4y. Let 3x+4y=a. This equation represents a straight line in the xy plane. The point M lies on the line, too. So we want the circle and line to have common points. That happens if-f the distance of the center from the line is less than or equal to the radius. From the formula of the distance between a point and a line we get |a|≤10. So the maximum value is 10. It is achieved when the line (for a=10) is tangent to the circle, that is at the point (6/5,8/5).
This can be done quickly using graphs or trigonometric parametric substitution
You were sent to attack the Graphs,not join them.
@@dadada6192 haha lol😂
@@gaussiit4855 I liked ur comment.
Can you do some permutations problems also? That will be fun
I solved this with analysis - get x from that circle without +/- because during f'(x) = 0 you will be turning it into square anyway and get the same results. You will get y = +/- 8/5 . I'm trying to find algebraic or geometric solutions, but when i know easier way my brain stops working.
x=2cost, y=2sin(t)
6cos(t)+8sin(t)=10(0.6cos(t)+0.8sin(t))
=10cos(t - theta)
cos(theta) = 0.6
sin(theta) = 0.8
max = 10 for angle t=theta+2kπ or t=-theta+2kπ where k in Z
Why also t=-theta+2kπ ?
@@MarcoMate87 generally cos is even so we have to include both series of solution but here we have
t-theta = 0 + 2kπ and you are right
Very elegant. But why run through the specific calculation after you have just proved the general case? Just substitute for a and b in the general formula. Are you trying to make the video longer? ;)
I appreciate this algebric trick but here is a derivation using trigonometric identites. As you mention in another response to a comment it boils down to maximizing (6costheta + 8sintheta), but this also reduces to maximizing cos(theta)+tan(phi)*sin(theta) where phi is the angle with tangent 4/3. Then this can be factored to read max(sin(phi+theta)) which is maximum for phi+theta=pi/2 and this means that tan(theta)=cot(phi)=3/4=x/y as in the condition you obtained algebrically and in some way justifies the inversion of the coefficients.
Sounds good!
This problem is not hard to solve by calculus, but that trick is really nice!!
If the sample provided for what would have been called the minimum value of the solution. I'm from Uzbekistan,
My solution (of course extreeeeeeeeeemely faster):
Sometimes I use algebra to solve a geometrical problem, this time I do the opposite.
x²+y²=4 is the equation of the circle centered on O of coordinates (0,0) and of radius 2.
3x+4y=K, where K is a real number, is the equation of a line. We're looking at the maximum value of K where the line intersects the circle.
We can see that when K rises, our lines moves upwards. If K is too low or too high, there are no points of the line that are also on the circle and when K rises inside the range where the line and the circle meet each other, the common points move to the north-east.
So the value of K that we're looking for is where the line is tangent to the circle in the north-east. But since the equation of the line is 3x+4y=K and since the radius is coming from O, we know that the radius belongs to the line of equation: y=4/3.x.
We report this into the equation of the circle: 25/9.x²=4 => x²=36/25 => x=6/5 (x is positive).
y=4/3.(6/5)=8/5.
Then 3x+4y=18/5+32/5=10.
Divide the equation by 4 both sides, then write it as (x/2)^2+(y/2)^2=1 then let x/2=sin theta so y/2=cos theta. Sub x=2sin theta and y=2 cos theta in the 2nd equation we get max(6sin theta +8cos theta). And, max and min value of a sin theta+ b cos theta is ± √(a^2+b^2) So √(6^2+8^2)=√100=10
Pretty interesting!
Why max and min is +or- √[a^2+b^2] ? Can we prove it by subsidiary angle?
@@鈴木悠真-n6z Yes, you can also do it by taking the derivative and equal it to 0
And then substitute the value of sin theta and cos theta
@@advaykumar9726 thank you!
How to analytically measure exact value of max?
This is new and nice sol.
Can anyone explain the derivative approach to this problem ?
put sin(z) = x/2 and cos(z) = y/2
then 3 sin(z) + 4 cos(z)
= 5 sin(z+ arc tan ( 4/3))
so max ( 3x+4y)
=2 max ( 10 sin(z+ arc tan ( 4/3))
= 10
This also can be solved using differentiation
How?
Most of the time you are taking a very lengthy approach to solve this kind of easy problems. I think taking an easy approach will make the students interesting.
Consider the family of parallel lines 3x + 4y = c; the idea is to maximize the value c.
there will be two parallels in this family tangential to the circle x2 + y2 = 4 and giving maximum value of c which will be 10.
👍
Nice, and that is gonna happen at x=1.2 and y=1.6 ... right ?
Thanks! Yes
I find your videos great, but it would be very helpful for me, if you tried to explain your plan before giving „random“ identities. I did not really get the sense of the identity before the end.
Thanks for helping me to get a better intuition on algebraic problem-solving!
Thanks for the tip!
Easy trig. substitution or by method of vectors, this q. can be solved. But your approach is formal & leaves no scope for any domain kind of a mistake! Congrats!!!
I did it orally using cauchy schwarz
1. Cauchy's inequality.
2. x= sin t, y= cos t.
3. Caculus.
Marvellous method of solving
Glad you think so!
Let's solve this problem using Lagrange multipliers.
Let F(x,y) = 3x + 4y, subject to the constraint G(x,y) = x^2 + y^2 - 4 = 0.
Fx = λGx & Fy = λGy, by theorem finds the extreme points, where λ is real and
Fx is the partial derivative of F wrt to x,
Fy is the partial derivative of F wrt to y,
Gx is the partial derivative of G wrt to x,
Gy is the partial derivative of G wrt to y.
Now Fx = 3 & Fy = 4 and Gx = 2x & Gy = 2y.
So, 3 = λ(2x) & 4 = λ(2y)
So, 9 = 4(λ^2)(x^2) & 16 = 4(λ^2)(y^2)
So, 25 = 4(λ^2)[x^2 + y^2]
So, 25 = 4(λ^2)[4], as x^2 + y^2 = 4
So, 25 = 16(λ^2)
So, λ = ±5/4 leading to (x,y) = (±6/5, ±8/5) as the extreme points, using 3 = λ(2x) & 4 = λ(2y), which results in the maximum of 3x + 4y being 10 when (x,y) = (6/5, 8/5), and the minimum of
3x + 4y being -10 when (x,y) = (-6/5, -8/5)
Imma solve this mah self.
Cool!
@@SyberMath And I didn't feel like it.
The problem can be easily solved using the method of Lagrange multiplier.
I love this comment section. Such love for Maths. I just used calculus to solve it. 🙂
Glad to hear that! 🥰
Hii mst syber !! As always amazing problems , until now i didn t watch your approach , i think my approach is more complicated than your s
I let y=k.x then let 3x + 4y = t
Now we must maximize t if you plug the substitution in the equations you get a system ..... finaly you get a function of k t(k)=......
You take the derivitive and you get one solution which is max
......... finaly you get t or (3x+4y) =10
Hi Tony! Your method is awesome! 🤩
@@SyberMath thank uu !!!!!!
Nice problem and solution. Objective fnc 3x+4y can be represented as A*sin(something). Max value is A which in our case is 10.
Absolutely beautiful and clever!!!
Give me more!!!
Thank you! Will do!