Question 1082936
I'll solve the 1/4 problem and then multiply by 4.
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*[illustration 98.JPG].
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{{{A=xy}}}
Define a new function which is the square of A,
{{{Z=A^2=x^2y^2}}}
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Make Z a function of one variable using the ellipse,
{{{4y^2=36-4x^2}}}
{{{y^2=(36-9x^2)/4}}}
Substituting,
{{{Z=(x^2(36-9x^2))/4}}}
Take the derivative of Z with respect to x and set it equal to zero.
{{{dZ/dx=-9x(x^2-2)=0}}}
{{{x=0}}} is the trivial solution so,
{{{x^2=2}}}
{{{x=sqrt(2)}}}
So then solving for {{{y^2}}},
{{{9(2)+4y^2=36}}}
{{{4y^2=18}}}
{{{y^2=9/2}}}
{{{y=3/sqrt(2)}}
{{{y=(3/2)sqrt(2)}}}
So then the maximum area would be,
{{{A[max]=4(sqrt(2))(3/sqrt(2))=12}}}
The width of the rectangle (x direction) would be,
{{{W=2sqrt(2)}}}
and the length of the rectangle (y direction) would be,
{{{L=2(3/2)sqrt(2)=3sqrt(2)}}}