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? 

Rather than using the constraint to find $x$ in terms of $y$ or vice versa, use implicit differentiation.
Differentiate the objective and set it equal to $0.$
$A = 4xy\\
\frac {dA}{dx} = 4y + 4x\frac {dy}{dx} = 0$
Differentiate the constraint.
$\frac {d}{dx} (4x^2 + 9y^2 = 3600)\\
8x + 18y\frac {dy}{dx} = 0\\
\frac {dy}{dx} = -\frac {4x}{9y}$ 
And now do your substitutions.
$-4x^2 + 9y^2 = 0\\
4x^2 + 9y^2 = 3600\\
y^2 = \frac {3600}{18}\\
|y| = \frac {20}{\sqrt 2}\\
|x| = \frac {30}{\sqrt 2}\\
4xy = 1200$
