Finding Local Maxima and Minima by Differentiation
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- Опубликовано: 8 фев 2025
- What else is differentiation good for? Well if we are looking at the graph of a function, differentiation makes it super easy to find where any local maxima and minima occur. This act in itself has many applications, but before we learn those, let's just learn how to find the maxima and minima!
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Sir 'Professor Dave Explains', i just wanna let you know how grateful and lucky i am. watching your calculus video is enlightening. i am from philippines, currently out of school. i am doing some review on my math subjects before re-enrolling again. i am also a self learner, and thanks to you because i am on a great progress with calculus (self taught). i never learned it from the school, i learned it on your calculus videos and some old books. you saved me.
im hoping that someday, if i happen to meet you in person, i have already achieved my dream of becoming a good mathematics teacher.
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From Philippines too❤️
nice one pre!
Thank you, professor Jesus.
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He looks like so!
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U like ellipses perchance?
He says "that one was pretty simple" just as my brain was about to explode.
Are you in AP calc?
It's very simple for us Asians 🤷
@@mariakhan6090 I'm part Asian too but I'm mainly English. If I pointed out I find programming easy, purely because I am English, people would think that was racist.
It was literally the easiest.
Noob
I’m not even in calc yet. I’ve exclusively watched these and can say I have a firm understanding of everything he says. Same for organic chem. Someone get him an award
Your great at what you do man! The reason these specialized videos don’t get many views is because only a select few need this type of information and your competing with khan academy etc. As more time goes on these type of specialized videos will have so many views and help so many people. I’ve already seen you covering more broad topics and they have a ton of views and are structured excellently for learning. I come to your videos for all my topics your amazing!!
woohoo thanks! yes i think that once i am finished with calculus i will find a way to advertise a little bit and then it will get a more appropriate viewership. but thanks for watching and spread the word!
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Calculus is magic, derivative is fucking wonderfulll!!!!
Done this lesson.
professor dave thnx for making this video i have my math exam in 3 days and this helped me a lot in learning maxima and minima thank you sir
Who is watching this video in 2024 ??
Also in 2025 January
Thanks I was struggling with this ..from quite a long time and finally got it. Thanks once again.
If there is one job in the world that can be this helpful and important in the world besides medicine it's teaching
a great source of required information.....Thank you sir...
Thank you sir for your dedication and for making this free! 🙏
Thank you Mr. Professor Dave, I would like to thank you for explaining this concept. Now I understand local maximum and minimum. May God Bless You and His Peace Be Upon Upon Always!
You had explained the whole concept in clear and precise way.... Very fantastic teacher u are
Thank you for the explanation Professor. Jesus Bless you always. After finished watching this video I immediately understand this material.
Why did I feel like a baby in the first place??
About to give up maths but thankyou professor ❤️🥺
Keep on enlightening inchoate mathematics tyros.
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Professor Jesus helped me pass my class, Thankyou
Sir excellent method of teaching
Fantastic explanation sir
Coming up to my second semester exams and this helps so much. What you do In 6 min takes my teacher 3 lessons
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Sir. I wanna you that you really helped me understand implicit and relative differenciation
Man I’ve always been a fan of your anti conspiracy stuff. I’ve just come into calculus and you are my hero
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Explain by profferser Dave is excellent
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Are you an actual professor you know everything
He answers this in one of his "Ask Professor Dave" videos.
Thanks for such a great explanation
Thanks from India
Thanks bro 👍
But how do we know which root at 3:14 was max or min without a graph?
check if after differentiation
if value more than 0 then minima
if value less yhan zero is maxima
You can do that by taking the derivative of the original function twice. This is the derivative of the derivative aka 2nd derivative.
1. You find critical pts (c) for x by setting 1st derivative of the original function to zero and solve for x. f(x)=0 solve for x.
2. Differentiate f(x) twice to get f''(x).
3. Plug in the critical values of x to evaluate f''(c).
4. If f''(c)
Thank you, Sir Dave!!!
Great review!
Very clear, very nice
I've passed my calculus classes already, but it's fun to watch it again
thank you so much, i was about to cry because i didnt know what to do =)
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professor dave im so sorry i strayed from your teachings i didnt know you had calc lessons
Yep 35 part playlist, check it out!
this one helps me a lot like im really stuck right now at my calculus. and im very afraid that my top grade goes down. this one really helps sir especially that we are now in new normal class❤❤
Thankyou
Nicely explained
Feels so crazy to see the connection from a precal concept lol.
For the 'comprehension' problem, why did you inset plus 2 and minus 2 into the equation afterwards?
because that's where the local maximum and minimum occur, and we want to get the exact maximum and minimum values
in the first example is y1 =1 and y2 =-3 ?
Thank u *Professor* --😍😍
Ur a blessed soul 🙏
Thanks for your great explanation.,.👍
Thanks bro 👍👍💯🔥🔥🧠
Thank you sir
Lmao i just stared at the video thumbnail for a bit, then i understood exactly what to do 😂😂😂. im just commenting to make sure you know u helped. 👍
3:10
If the curve is inverted W.R.T x-axis. Then also you will get 2 extrema at x=0 and x=2
But how to tell if that's maxima or minima
Subs value in f(x).. smaller value is minima and maximum value is maxima
thank you so much
I have a question. I know this is a bit late but why did we plug back in 2 and -2 in the f(x) function? Is it because to find the (x,y) values? Please let me know and thank you!
yup to locate the maximum and minimum points :D
:3@@rxsvie
You know what they say, "God is always there for you even when your math teacher sucks"
My precalculus teacher sucks. I have to teach the whole class during homework time
@@isaacfandakly lmao way to goo!
This is the entire economics profession
Bro is the 💎. And is found in the caves of RUclips i am so lucky to find him
Can you tell me in which grade it is studied
Depends, in India we learn this in 11th
What's a English speaking nice and fantastic understanding
What?
yes
How do you find the maximum number of zero crossings?
Degree of equation= maximum number of zero crossing
Better than uni tutor x100
I just really love intro
Can u plz give a simple daily life application of maxima and minina! From many videos i noticed that it helps in economics to maximize and ,in efficienting packing etc..But none of them explains how it is done...well,after finding local maximum,how can it help to maximize profit?!!plz help to figure it out
That's not a "simple daily life application." There's a reason why people study these things for years. However, to answer your question: for this, you'd have to have your revenue and cost function. Subtract the C from R, and from there, just find the local maximum and it should tell you how much you need to set the price for the product. If you want to take it a step further, you can minimize the average cost.
Here's a simple example of optimization, that you could easily understand its application to real life. Suppose we need to transport cargo from an island to a warehouse. Consider a straight shoreline that runs east-west, and the warehouse is L=100 km east of the island, also on the shore. Assume the road runs directly along the shore and any inland travel distance to get to the road is negligible. The island is d=50 km south of the shore. We need to determine where along the shoreline to build our dock for unloading the ship, that will minimize the amount of fuel consumed.
Suppose a truck takes 1 liter of fuel to drive 1 km (a rate we can call T) and suppose a ship takes 5 liters of fuel to transport the same cargo a distance of 1 km (a rate we'll call S). Define the origin at the point on the shore directly across the water from the island, and define distance x as the position along this shore where the dock could be located. Assume only a 1-way trip for simplicity.
A: find an expression for the fuel consumed by the truck, as a function of dock location x
B: find an expression for the fuel consumed by the ship, as a function of dock location x
C: find a function f(x) for the total fuel consumed, by combining expressions from parts A and B. Answer: f(x) = T*(L-x) + S*sqrt(x^2 + d^2)
D: Take the derivative of this function.
E: Set this derivative equal to zero, and solve for the optimal value of x that minimizes fuel consumption. Answer: x=(d*T)/sqrt(S^2 - T^2), which happens at x=10.2 km.
Thanks 👍
You have not made a video about integration by parial fraction decomposition
You are wonderful
You did put the x values once in first derivative and then in the last problem you did put the value in the original function? Why
You find the x values for which the slope is 0 using the derivative and plug those x values into the original function to get the y value.
Also make a video about differential equation
Professor Dave question(-tsquared+4t+4)
That doesn’t factor
I had a math question, how can I send it to you?
U r good
Any link where I can find a whole. Playlist on applications of differentiation and integration?
Check my calculus playlist!
thanks a lot!!!!
can someone explain to me why at 3:50 2x^2 disappears?
Kind of late but in case anyone else is wondering about that : x^2+1 -x *2x = x^2+1 -2x^2 = x^2-2x^2+1 = -x^2+1
4 years ago and still helped me passed calculus1
Hi sir! Can help me to solve this problem? Please 🙏🙏🙏 It is so hard for me ☹️
Problem:
A rectangular box with a square base is inscribed in a hemisphere of radius R. Find the maximum volume of the box.
Define capital X to be the half-width of the rectangular box. This means that the area of the base of the box will be (2*X)^2. You'll see why we choose to work with the half-width instead of the full width, very soon. We know that X can at a minimum equal zero, and at a maximum equal R, so we will only be interested in critical points within this range.
Create an equation for the hemispherical dome, in terms of horizontal positions x and y.
Equation of a sphere:
x^2 + y^2 + z^2 = R^2
Equation of the positive half of the sphere:
z = sqrt(R^2 - x^2 - y^2)
Your box will have a corner at a point where y=x, and also equals capital X. Use the previous equation to determine H, the height of the box.
H = sqrt(R^2 - 2*X^2)
Now define V for the volume of the box:
V = H*(2*X)^2
Simplify:
V = 4*H*X^2
Plug in H, and now we have our objective equation only in terms of the variable X:
V = 4*X^2*sqrt(R^2 - 2*X^2)
Take the derivative dV/dX, which we will set to zero to look for the critical value of X where V is maximum. Along the way, you'll be using the product rule, the chain rule, and the power rule. I'll skip the steps and jump right to the result.
dV/dX = (8*X*(R^2 - 3*X^2))/sqrt(R^2 - 2*X^2)
We are interested in where dV/dX = 0, but we also need to watch out for locations where the denominator is also zero. Because if these coincide, we have a hole in the function instead of a zero.
Locations where numerator equals zero:
Trivial answer at X = 0, due to 8*X equaling zero at this point.
More interesting answer:
R^2 - 3*X^2 = 0
Solve for X:
X = R/sqrt(3)
Denominator of zero:
R^2 - 2*X^2 = 0
Solve for X:
X = R/sqrt(2)
Since these do not coincide, it is acceptable to conclude that X = 0 and X = R/sqrt(3) are our critical points where dV/dX = 0. What ends up happening at X=R/sqrt(2), is that you have the maximum possible square base that can be inscribed in the circle, and its height is zero.
We know X=0 is not the maximum, because the box would have no thickness and thus no volume.
Therefore, we conclude the critical point X = R/sqrt(3) is the location where the volume is maximized. This means that the box's base dimensions are each equal to 2*R/sqrt(3), and the box's height is R/sqrt(3).
@@carultch please help me I have a request to you. How to find maxima and maxima of the type:
y=k/f(x)
Where f(x)=ax²+bx+c or,|x-a|+|x-b|
@@simon-gh1pt For y=k/(a*x^2 + b*x + c), start by finding the poles of the function (i.e. places where the denominator equals zero). These aren't necessarily the minima/maxima, but they are important points to know. These are locations where there could be a vertical asymptote, meaning the function approaches either positive or negative infinity, or both from opposite sides. This would mean the maximum or minimum is unlimited, as immediately adjacent to the pole, would be an extremely large output. That is, unless a pole coincides with a zero in the numerator, at which point there will be a removable singularity. If there are no real solutions to the denominator equaling zero, it is in your favor, because this means you will have no singularities.
Since this is a quadratic expression, we can use the quadratic formula to find where the denominator equals zero. To the tune of pop goes the weasel: "x is equal to negative b, plus and minus the square root; of b squared minus 4*a*c, all over 2*a."
Next, we are interested in locations where dy/dx = 0, which means the function will be locally flat, and reach a turning point. Take the derivative, and set it equal to zero.
Rewrite as y = k/f(x)
dy/df = k*1/f^2
dy/dx = dy/df * df/dx = k/f(x)^2 * f'(x)
Take derivative f'(x) = 2*a*x + b, per the power rule.
Reconstruct:
dy/dx = k*(2*a*x + b)/(a*x^2 + b*x + c)^2
The only way that this function can equal zero, is when 2*a*x + b = 0. Solve for x to be x = -b/(2*a).
This will mean that the extreme point will occur at x = -b/(2*a).
Take the second derivative and evaluate at x=-b/(2*a), to determine if this is a local minimum (positive 2nd derivative) or a local maximum. This will be the only instance where there is a local extreme point that isn't part of a singularity. The second derivative evaluated at this point equals -(32*k*a^3)/(4*a*c - b^2)^2. When --k*a^3 is positive, you'll have a local minimum. When -k*a^3 is negative, you'll have a local maximum.
Note that in the event that b^2 - 4*a*c = 0, then you will also have a denominator equal to zero in the original function at the same point where the derivative equals zero. This means it is inconclusive to call it a local minimum or maximum, because this will also coincide with a vertical asymptote.
Ideally, b^2-4*a*c
@@simon-gh1pt
As for y=k/(|x - a| + |x - b|), this one has another unique challenge in that the denominator function is not differentiable. There will be kinks/cusps where the interior of the absolute value bars equals zero. This is what you end up with as the derivative:
dy/dx = (k*(sgn(a - x) + sgn(b - x)))/(abs(a - x) + abs(b - x))^2
where sgn refers to a function that looks at the sign of the input, and either returns +1 or -1. It returns 0 by convention when the input equals zero.
You will also end up with a flat-line between x=a and x=b, so there is an entire range of possible answers for the local maximum or minimum.
how to find the maxima and minima of a damping signal as it has more than 1 point with m=0?
Thank you, this was really helpful 😁
If there are more then 1 local maxima/minima can we find all of them
The intro ❤❤
How do you find the function when only given the maxima and minima?
You can't. There are an infinite number of functions that have the same minima and maxima.
I can just say love u sir..........
Very good
perfecto
Hi Prof, @3.07 by looking the graph i can say local min & max are -2,+2 but the solution you have is 0,2. Im bit confused
Dave I feel like f prime of X isn't right on 3.46
hello sir,I want to receive lectures of calculus from you. how can I get that? sir.