Inscribing a rectangle in an ellipse (optimization)

[CharismaticBarber]

So the problem goes as follows:

Find the area of the largest rectangle that can be inscribed in the ellipse (x^2/a^2) + (y^2/b^2) = 1 and verify that it is the absolute maximum area.

It gives the answer as 2ab

I've gotten as far as solving for y and plug it into A=4xy, which is y=4xbsqrt(1-(x^2/a^2) but I'm not sure what to do after this.

If you could show all the work in each step that'd be great because I have a test on this tomorrow and am almost totally lost hahaha. Thanks!

 

[Ishuda]

So the problem goes as follows:

Find the area of the largest rectangle that can be inscribed in the ellipse (x^2/a^2) + (y^2/b^2) = 1 and verify that it is the absolute maximum area.

It gives the answer as 2ab

I've gotten as far as solving for y and plug it into A=4xy, which is y=4xbsqrt(1-(x^2/a^2) but I'm not sure what to do after this.

If you could show all the work in each step that'd be great because I have a test on this tomorrow and am almost totally lost hahaha. Thanks!

First you need to justify why you can take the sides of the rectangle parallel to the x and y axis. After that, you have (with a minor correction of changing the y to an A which I'm pretty sure you meant)
A = 4 b x $\displaystyle \sqrt{1 - (\frac{x}{a})^2}$

Since this is asked in the Calculus forum what do you know about the derivative of a function when the function is at a maximum or minimum? How do you know whether it is a maximum or minimum or just a point of inflection?

 

[CharismaticBarber]

First you need to justify why you can take the sides of the rectangle parallel to the x and y axis. After that, you have (with a minor correction of changing the y to an A which I'm pretty sure you meant)
A = 4 b x $\displaystyle \sqrt{1 - (\frac{x}{a})^2}$

Since this is asked in the Calculus forum what do you know about the derivative of a function when the function is at a maximum or minimum? How do you know whether it is a maximum or minimum or just a point of inflection?

I know that you must find dA/dx and then find where x=0 for that derivative, then plug that value into A, I'm just having trouble finding the derivative.

 

[Ishuda]

I know that you must find dA/dx and then find where x=0 for that derivative, then plug that value into A, I'm just having trouble finding the derivative.

Does it help to write it as
A = 4 b x $\displaystyle [ 1 - (\frac{x}{a})^2 ]^{\frac{1}{2}}$

If not, I believe you need some review. In particular, for this problem, the product rule
(fg)' = f' g + f g'
and the power rule
f(x) = [ g(x) ]ⁿ
then
f'(x) = n [ g(x) ]^n-1 g'(x)

If you are having problems after this, please show what you have done so far so we can see specifically where you mght need help.

 

[HallsofIvy]

I would NOT solve for y in terms of x (or vice-versa). I would treat this as "maximize A(x,y)= 4xy with the constraint $\displaystyle g(x, y)= \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$". $\displaystyle \nabla A= 4y\vec{i}+ 4x\vec{j}$ and $\displaystyle \nabla g= \frac{2x}{a^2}\vec{i}+ \frac{2y}{b^2}$. At a max or min for g, under that constraint, those two vectors must be parallel- one is a multiple of the other. Taking that multiple to be $\displaystyle \lambda$ (that's the "Lagrange multiplier), we must have $\displaystyle 4y= \lambda\frac{2x}{a^2},$, $\displaystyle 4x= \lambda\frac{2y}{b^2}$. Dividing the first equation by the second eliminates $\displaystyle \lambda$ and we have, $\displaystyle \frac{y}{x}= \frac{b^2}{a^2}\frac{x}{y}$ so that $\displaystyle y^2= \frac{b^2}{a^2}x^2$.

If $\displaystyle y= \frac{b}{a}x$, then the constraint gives $\displaystyle \frac{x^2}{a^2}+ \frac{b^2}{a^2}\frac{x^2}{b^2}= \frac{2x^2}{a^2}= 1$ so that $\displaystyle x= \frac{a}{\sqrt{2}}$. Then $\displaystyle y= \frac{b}{a}\frac{a}{\sqrt{2}}= \frac{b}{\sqrt{2}}$ and the area is $\displaystyle xy= \frac{ab}{2}$.

 

[Ishuda]

... and the area is $\displaystyle xy= \frac{ab}{2}$.

Shouldn't that be the first quadrant area? If x were -a/2 then the fourth quadrant area would be ab/2, etc. and the total area would be 2 ab