SOLUTION: Given that x and y are positive real numbers such that x^2-2x+4y^2=0, btain the maximum value of the product xy.

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Question 1104622: Given that x and y are positive real numbers such that x^2-2x+4y^2=0, btain the maximum value of the product xy.
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Solve the quadratic for either x or y and substitute into the product xy to get the product as the function of a single variable. Then differentiate that product and find where the derivative is zero.

x%5E2-2x%2B4y%5E2+=+0
4y%5E2+=+2x-x%5E2
y%5E2+=+%282x-x%5E2%29%2F4
y+=+%28sqrt%282x-x%5E2%29%29%2F2 [since the problem says x and y are both positive]

Then the product xy is
x%2A%28sqrt%282x-x%5E2%29%29%2F2+=+sqrt%282x%5E3-x%5E4%29%2F2 [note: when I have to take a derivative like this, I find it easier to move the "x" inside the radical, so that I don't have to use the product rule when doing the differentiation.]

The derivative is
%286x%5E2-4x%5E3%29%2F%284%2Asqrt%282x%5E3-x%5E4%29%29

The derivative is zero when
6x%5E2-4x%5E3+=+0
3+-+2x+=+0
x+=+3%2F2

When x = 3/2,
y+=+sqrt%283-9%2F4%29%2F2+=+sqrt%283%2F4%29%2F2+=+sqrt%283%29%2F4

And at that point the product xy is
%283%2F2%29%2A%28sqrt%283%29%2F4%29+=+3%2Asqrt%283%29%2F8

The maximum value of xy, given x%5E2-2x%2B4y%5E2=0, is 3%2Asqrt%283%29%2F8