SOLUTION: maximize q=5xy^2, where x and y are positive numbers such that x+y^2=8

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Question 1051078: maximize q=5xy^2, where x and y are positive numbers such that x+y^2=8
Found 2 solutions by Fombitz, ikleyn:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
x=8-y%5E2
Substitute into q equation,
q=5%288-y%5E2%29y%5E2
Now q is a function of one variable, y.
To find the maximum, differentiate with respect to y and set the derivative equal to zero.
dq%2Fdy=5%288-y%5E2%29%282y%29%2B5%28y%5E2%29%28-2y%29
dq%2Fdy=80y-10y%5E3-10y%5E3
dq%2Fdy=80y-20y%5E3
So,
80y=20y%5E3
y%5E2=4
Remember y is positive,
y=2
So then,
x%2B2%5E2=8
x%2B4=8
x=4
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q%5Bmax%5D=5%284%29%284%29=80

Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
maximize q=5xy^2, where x and y are positive numbers such that x+y^2=8
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The solution at the Algebra level, with no calculus. 


x=8-y%5E2

Substitute into q equation,

q=5%288-y%5E2%29y%5E2

Now q is a function of one variable, y.

q = 40y%5E2+-+5y%5E4.

"Complete the square" :

q = -5%28y%5E4+-+8y%5E2%29 = -5%2A%28%28y%5E2+-+4%29%5E2+-+16%29 = -5%28y%5E2-4%29%5E2 + 80.

It shows that q is maximal at  y%5E2 = 4,  or  y = +/-2.

According to the condition, only positive "y" are considered, so the solution is y=2.

Then x = 4.

The minimum of q is 80.