Calculus 1: Given x+2y=4, find the max of sqrt(x)+sqrt(y)

How do we find the maximum of a function when we are given a constrain equation. Here we have x+2y=4 and how do we find the maximum of sqrt(x)+sqrt(y)?
To see how to do this without using calculus, check out this video: • How Cauchy would find ... (Cauchy Inequality)
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[@vascomanteigas9433]

Using Lagrange Multipliers are more straightforward.

[@taldoedu7615]

You also can use RMS-AM inequality:
Set *E = √x + √y*
-> E = 1/2 √x + 1/2 √x + √y
-> E/3 = [1/2 √x + 1/2 √x + √y]/3 ≤ √[(x/4 + x/4 + y)/3]
-> E/3 ≤ √(x + 2y)/6 = √(4/6) = √(2/3)
-> E ≤ 3.√(2/3) = √6
=> *√x + √y ≤ √6*

[@maxgoldman8903]

Another method: We can find the maximum using the AM-GM inequality.
(√ x+√ y)² = x+y+2•√ (xy)
= x+y+2•√ (x/2)√ (2y) ≤ x+y+(x/2+2y)
= 3/2x+3y = 3/2•(x+2y) = 3/2•4 = 6.
Max (√ x+√ y) = √ 6.

[@jubinsoni4694]

When the question was shared earlier, I used the method suggested in this video, thanks for sharing

[@AtheFbEast]

Cool , i would really like to see algebra answer

[@bprpcalculusbasics]

Here ruclips.net/video/YSJzpQQSOZw/видео.htmlsi=70Vmlw8kD58DW79m

[@AtheFbEast]

@@bprpcalculusbasics thx

[@iAzazelHD]

Lagrange multipliers when :(

[@timothymoorhouse1200]

Could you tell me where to find the Calculus 1, 2 and 3 curricula that you teach for? I'm curious after watching quite a number of your videos.

[@economicist2011]

3:40 You can save yourself a lot of irritation by just setting S'(y) = 0 at this point... no need to go finding common denominators or anything...
0 = -1/sqrt(4-2y) + 1/[2*sqrt(y)]
1/sqrt(4-2y) = 1/[2*sqrt(y)]
sqrt(4-2y)=2*sqrt(y)
4-2y = 4y
6y = 4
y = 2/3 --> x = 8/3

[@RikiFaridoke]

It is easy enough, surely we can solving it via derivative aplication method.

[@JimmyCerra]

I used a weird method:
1. D[x+2y=4] = dx + 2dy = 0, or dx = -2dy
2. D[S = sqrt(x) + sqrt(y)] = S' = 0
S' = dx/(2*sqrt(x)) + dy/(2*sqrt(y)) = 0
dx + dy * sqrt(x/y) = 0
Undefined at (from 2 and x+2y=4):
x != 0 so y != 2, y != 0, so x != 4
Combining 1 & 2:
3. (sqrt(x/y) - 2) * dy = 0
sqrt(x/y) - 2 = 0
x/y = 4, x = 4y
4y + 2y = 4, y = 2/3, x = 8/3
Critical points:
S(0,2) = sqrt(2), S(4,0) = sqrt(4) = 2, S(2/3,8/3) = sqrt(6)
So max = sqrt(6).

[@user-yn5bk2pe9o]

How the logic works?
Why D[x+2y=4] works
Or u mean using partial derivative?

[@mayureshious]

What if we use 2nd derivative test for maxima? If we consider ds/dy =0 then I think we get only one critical point.

[@saravanarajeswaran2626]

yeah i did that and i got the same result root6

[@kaewoomusic]

this is more complicated than the regular method 😭

[@Metaverse-d9f]

you wasted 2min. doing the table!

[@Lolwutdesu9000]

Why do we ignore the negative roots of x and y?