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.
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
It's just a convention. The square root symbol used on a positive real number means the positive square root. If we need both square roots, we use the ± symbol in front of it (like in the quadratic formula, for example). Also in this case it's obvious that the negative roots wouldn't give us the max value.
Using Lagrange Multipliers are more straightforward.
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.
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*
When the question was shared earlier, I used the method suggested in this video, thanks for sharing
Cool , i would really like to see algebra answer
Here ua-cam.com/video/YSJzpQQSOZw/v-deo.htmlsi=70Vmlw8kD58DW79m
@@bprpcalculusbasics thx
Lagrange multipliers when :(
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.
It is easy enough, surely we can solving it via derivative aplication method.
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
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).
How the logic works?
Why D[x+2y=4] works
Or u mean using partial derivative?
What if we use 2nd derivative test for maxima? If we consider ds/dy =0 then I think we get only one critical point.
yeah i did that and i got the same result root6
this is more complicated than the regular method 😭
you wasted 2min. doing the table!
Why do we ignore the negative roots of x and y?
It's just a convention. The square root symbol used on a positive real number means the positive square root. If we need both square roots, we use the ± symbol in front of it (like in the quadratic formula, for example).
Also in this case it's obvious that the negative roots wouldn't give us the max value.