Area of the largest rectangle with sides parallel to $x$- and $y$-axes that can be inscribed in the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$

Question

First off, I simplified the equation of the ellipse to get it to

$$9x^2 + 16y^2 = 144$$

And then did further simplification to get it in terms of:

$$y= \bigg(9 - \frac{9 x^2}{16}\bigg)^{\frac{1}{2}}$$

So I then plug in $y$ to the equation of the area of a rectangle:

$$A = xy = x(9 - \frac{9x^2}{16})^{\frac{1}{2}}$$

And then I proceeded to differentiate the expression with respect to $x$ and derived:

$$\frac{dA}{dx} = (9-\frac{9x^2}{16})^{\frac12} + 9x^2\cdot(16(9-\frac{9x^2}{16})^{-1})$$

After I attempted to find the critical points of $\frac{dA}{dx}$, I multiplied both sides of the equation by:

$$16(9-\frac{9x^2}{16})^{-1}$$

And then got:

$$144 - 9x^2 + 9x^2 = 0$$

And this is where I knew all this work had been done wrong, as my relation of the area of a rectangle and the ellipse was wrong. Can anyone Show me the right way to start of the problem to solve it?

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Jul 03 '18 at 09:44
You didn't take your derivative correctly. The second part of your product rule is pretty messed up. The first thing is that the last exponent should be $-1/2$ not $-1.$ — B. Goddard, Jul 03 '18 at 11:00
Possible duplicate of Find the area of largest rectangle that can be inscribed in an ellipse — Jyrki Lahtonen, Jul 03 '18 at 11:30
Search time approx six seconds. You may prefer to use this as a dupe target instead. I don't remember what swayed my pick in 2015. — Jyrki Lahtonen, Jul 03 '18 at 11:31

score 2 · Answer 1 · answered Jul 03 '18 at 13:44

Your differentiation wasn't correct......

$A = xy = x(9 - \frac{9x^2}{16})^{\frac12}$

$\frac{dA}{dx} = (9 - \frac{9x^2}{16})^{\frac12}+\frac{x}{2}(9 - \frac{9x^2}{16})^{-\frac{1}{2}}\cdot -\frac{9x}{8}$

$\frac{dA}{dx} = (9 - \frac{9x^2}{16})^{\frac12}-\frac{9x^2}{16}(9 - \frac{9x^2}{16})^{-\frac{1}{2}}$

Area of the largest rectangle with sides parallel to $x$- and $y$-axes that can be inscribed in the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$

1 Answers1