Doug C. answered • 10/10/16

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Never mind. You are on the right track. Could have used the product rule too.

Mathalina S.

asked • 10/10/16It says to eliminate one variable so I eliminated x.

A=xy

L^{2}=x^{2}+y^{2}

x=(L^{2}-y^{2})^{1/2}

A=y*(L^{2}-y^{2})^{1/2}

=(L^{2}y^{2}-y^{4})^{1/2}

Derivative =1/2(L^{2}y^{2}-y^{4})^{-1/2} *L^{2}2y-4y^{3}

I don't know where to go from here

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Doug C. answered • 10/10/16

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Never mind. You are on the right track. Could have used the product rule too.

Peter G. answered • 10/10/16

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You're on the right track.

Now factor out a y from the numerator and a y from the denominator; these cancel. Cancel a 2 in the numerator and denominator. You are left with

0 = (L^{2} - 2y^{2})/√(L^{2}-y^{2}), so

0 = L^{2}-2y^{2}

y = L/√2.

This shows that area is maximized in a square, i.e. when x = y. At this point, A = L^{2}/2.

It complicated things when you moved the y inside the radical. I would have left it outside, and used the product rule.

I hope that helps.

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