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? 
 A: There is no need of any complicated algebra. Following is a geometric way to get the answer. One advantage of this approach is you don't need to assume the largest rectangle is axis aligned with the ellipse.
Given any circle, it is well known the largest quadrilateral inscribed in it is a square. Furthermore, the area of the square is $\frac{2}{\pi}$ of that of the circle.
Given an ellipse of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ and any rectangle inscribed in it. Under the linear transform $(x,y) \mapsto \left(\frac{x}{a},\frac{y}{b}\right)$, the ellipse and rectangle get mapped to a circle and quadrilateral. Since under such a linear transform, the ratio of area of different geometric shapes is invariant, we can deduce the area of the rectangle is at most $\frac{2}{\pi}\times \pi a b = 2ab$.
For the ellipse at hand, $a = 30$ and $b = 20$ and it is obvious how to find a rectangle which will get mapped to a square. This means the area of largest rectangle inscribed in that ellipse is $2ab = 2(30)(20) = 1200$.
A: 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.
$-4x^2 + 9y^2 = 0\\
4x^2 + 9y^2 = 3600\\
y^2 = \frac {3600}{18}\\
|y| = \frac {20}{\sqrt 2}\\
|x| = \frac {30}{\sqrt 2}\\
4xy = 1200$
A: Alternative way by Lagrange's multipliers


*

*$4y=8\lambda x$

*$4x=18\lambda y$


since $\lambda=0$, $y=0$, $x=0$ don't lead to any solution we can divide and obtain


*

*$\frac y x = \frac 49\frac x y \implies 9y^2=4x^2 \\\implies 8x^2=3600 \implies x^2=450\implies x=\pm15\sqrt 2\implies y=\pm 10 \sqrt 2$


and then
$$A=4|x||y|=1200$$
A: 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 $\,2x=3y\,$.
A: You seem to be going in the right direction. There is a mistake somewhere, but your algebra is a bit convoluted, so I cannot find it exactly. Here is what I got following your procedure:
Solve $4x^2+9y^2=3600$ for $y$ to get $$y=\frac23\sqrt{(30-x)(x+30)}.$$ Then, we want to maximize $$A=4xy=\frac{8x}{3}\sqrt{(30-x)(x+30)}.$$ Taking the first order condition we get $$\frac{d}{dx}\left(\frac{8x}{3}\sqrt{(30-x)(x+30)}\right)=0\;\Rightarrow\;-\frac{16}{3}\frac{x^2-450}{\sqrt{900-x²}}=0\;\Rightarrow\;x=15\sqrt{2}.$$
Plugging this back into the formula for $A$ we get
$$A=\frac{8*15\sqrt{2}}{3}\sqrt{(30-15\sqrt{2})(15\sqrt{2}+30)}=1200.$$
