<s>[INST] Calculating the largest possible area of a rectangle inscribed in an ellipse So i got the equation $4x^2 + 9y^2 = 3600$
What i've done so far is:
 $A= (2x)(2y) = 4xy$
Then I find the expression of $y$
$9y^2= 3600 -4x^2$
$y = \pm \sqrt{3600 -4x^2 / 9} = 2/3(\sqrt {900 - x^2} \quad 2/3(900 -x^2)^{1/2}$
Then i set 
$A = 4x(2/3(900 -x^2)^1/2 = (8/3)x(900 -x^2)^1/2$
Taking the derivative
$A'(x) = 8/3(900 -x^2)^{1/2} + (8/3)x(1/2)(900 -x^2)^{-1/2}(-2x)
= (2400 - (16/3)x^2)/(\sqrt{900-x^2})$
Set the $A' = 0$
$2400 - (16/3)x^2 = 0$
$(16/3)x^2 = 2400$
$(16/3)x = \sqrt{2400} = (20\sqrt{6}) / 3$
$x = (5\sqrt{6})/12$
Then i put the value of x in the equation and get
$A = (8/3)((5 \sqrt 2)/12)(900 - ((5\sqrt 2)/12)^2)^{1/2} = 81.6...$
Is this right or? 
 [/INST] Alt. hint  -   by AM-GM:
$$3600 = (2x)^2 + (3y)^2 \ge 2 \sqrt{(2x)^2 \cdot (3y)^2} = 12 \,|x|\,|y| = 3\,A$$
Equality holds iff $\,2x=3y\,$.
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