# Derivative Optimization Problem [duplicate]

I need help with finding the area of the largest rectangle in an ellipse from $y^2 + (x^2)/4 = 1$.

I got it to y = $\sqrt{ 1 - (x^2)/4}$ but then I don't really know what to do, please help.

## marked as duplicate by almagest, amd, Henrik, user147263, ShaileshMay 21 '16 at 0:08

• – lab bhattacharjee May 20 '16 at 16:51
• Didn't quite understand how they solved it, but thank you anyway – Tobias May 20 '16 at 17:02

I guess the area is defined as $x\times y$. So far, you got $y=\sqrt{1-x^2/4}$, therefore

$$A(x)=x\times\sqrt{1-x^2/4} \quad \text{for}\quad -2 \leq x \leq 2$$

Negative values do not make sense, so these might be ruled out. Also, the area would be $0$ for $x=2$ and $x=0$, so $0 < x < 2$.

Now you have to find the maximum of this function by differentiating.

$$\frac{dA}{dx}=-\frac{x^2-2}{2\sqrt{1-x^2/4}}$$

The slope at the highest point is $0$, so

$$0=-\frac{x^2-2}{2\sqrt{1-x^2/4}}$$

A fraction is only $0$, if the numerator is $0$:

$$0=x^2-2$$

Finally, $$x = \sqrt{2}$$

With this solution, you can find the final maximum area of the rectangle.

• Where did x^2 - 2 come from? – Tobias May 22 '16 at 13:14
• @Tobias By differentiating $A(x)$ and simplifying the result (in this case there are many variations that may not look the same). I have omitted the step by step solution, because it would take up too much space. – qwertz May 22 '16 at 18:36
• That makes sense! Thank you for the help. I think I have solved it now. – Tobias May 23 '16 at 11:23