Question 1104622
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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.<br>
{{{x^2-2x+4y^2 = 0}}}
{{{4y^2 = 2x-x^2}}}
{{{y^2 = (2x-x^2)/4}}}
{{{y = (sqrt(2x-x^2))/2}}}  [since the problem says x and y are both positive]<br>
Then the product xy is
{{{x*(sqrt(2x-x^2))/2 = sqrt(2x^3-x^4)/2}}}  [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.]<br>
The derivative is
{{{(6x^2-4x^3)/(4*sqrt(2x^3-x^4))}}}<br>
The derivative is zero when
{{{6x^2-4x^3 = 0}}}
{{{3 - 2x = 0}}}
{{{x = 3/2}}}<br>
When x = 3/2,
{{{y = sqrt(3-9/4)/2 = sqrt(3/4)/2 = sqrt(3)/4}}}<br>
And at that point the product xy is
{{{(3/2)*(sqrt(3)/4) = 3*sqrt(3)/8}}}<br>
The maximum value of xy, given {{{x^2-2x+4y^2=0}}}, is {{{3*sqrt(3)/8}}}<br>