Rectangle Inscribed in Ellipse

[danizh]

Problem: Find the dimensions of the rectangle of maximum area that can be inscribed in the ellipse x^2/16 + y^2/9 = 1.

I tried isolating "y," so that I can plug it into the rectangle area equation (A = xy).
Now I get:

y = [3 root(16-x^2)] / 4
I tried plugging this into A = xy, and then taking the derivative to find the maximum area, but I can't seem to get the correct answer. Any help would be great.

 

[wjm11]

Problem: Find the dimensions of the rectangle of maximum area that can be inscribed in the ellipse x^2/16 + y^2/9 = 1.

I tried isolating "y," so that I can plug it into the rectangle area equation (A = xy).
Now I get:

y = [3 root(16-x^2)] / 4

x^2/16 + y^2/9 = 1
(y^2)/9 = 1- (x^2)/16
y^2 = 9[1- (x^2)/16]
y = (9[1- (x^2)/16])^(.5)
y = 3[1- (x^2)/16]^(.5)

 

[danizh]

But I still can't get the correct answer.

A = (x)(y)
A = 3x [1- (x^2)/16]^(.5)

Now I can take the derivative:
A' = 1.5x [1 - (3x^2)/16]^(-1/2) x (-3x/8)
But A' = 0
so 0 = 1.5x
Therefore, according to this, x = 0.

This is obviously incorrect, so any help would be great as to what I'm doing wrong.

 

[stapel]

Check your work on your derivative of A. You might want to rewrite as:

. . . . .\(\displaystyle \Large{A\mbox{ }=\mbox{ }\left(\frac{3}{4}x\right)\mbox{ }\sqrt{16\mbox{ }-\mbox{ }x^2}}\)

Then apply the Product Rule.

Eliz.

 

[danizh]

By using the product rule, I am getting:
0 = - 16 + x^2 + 256x^4 - 32x^6 + x^8
I just can't seem to be getting this question right. :x

 

[stapel]

What happened to the square root?

Eliz.

 

[soroban]

Hello, danizh!

Just what is The Correct Answer?
. . Are you off by a factor of two?

Find the dimensions of the rectangle of maximum area
that can be inscribed in the ellipse: \(\displaystyle \frac{x^2}{16}\,+\,\frac{y^2}{9}\:=\:1\)

Your game plan is excellent.

                      |3
                   *  *  *
              *-------+-------* 
             *|       |       |*
            * |       |      y| *
      - - - * + - - - + - - - + * - - -
            * |       |      y| *4
             *|   x   |   x   |*
              *-------+-------*
                   *  *  *
                      |

Note that the area of the rectangle is: .\(\displaystyle A\:=\2x)(2y)\:=\:\)4xy.


Since \(\displaystyle y\:=\:\frac{3}{4}\sqrt{16\,-\,x^2}\), the area function is: .\(\displaystyle A\;=\;4x\cdot\frac{3}{4}\sqrt{16-x^2}\;=\;3x(16-x^2)^{\frac{1}{2}}\)

Differentiate: .\(\displaystyle A'\;=\;3x\cdot\frac{1}{2}(16-x^2)^{-\frac{1}{2}}\cdot(-2x)\:+\:3(16-x^2)^{\frac{1}{2}}\)

And we have: .\(\displaystyle \frac{-3x^2}{\sqrt{16-x^2}}\,+\,3\sqrt{16-x^2}\;=\;0\)

Multiply by \(\displaystyle \sqrt{16-x^2}:\;\;-3x^2\,+\,3(16\,-\,x^2)\:=\:0\)

. . which simplifies to: .\(\displaystyle x^2\,=\,8\;\;\Rightarrow\;\;x\,=\,2\sqrt{2}\) . . . then: .\(\displaystyle y\,=\,\frac{3}{2}\sqrt{2}\)


Therefore: .\(\displaystyle \text{Length} = 4\sqrt{2},\;\;\text{Width} = 3\sqrt{2}\)