# Max square/rec area under the curve

• Jul 22nd 2009, 04:40 PM
mant1s
Max square/rec area under the curve
Hi everyone, I've been working on this optimization problem for awhile now. I guess I don't understand the steps. Here's my work.

The Problem
Code:

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 4 - x^2.  What are the dimensions of such a rectangle with the greatest possible area?
My work

Code:

Define variables... A = 2xy y = 4- x^2 setup problem: A(x) = 2x ( 4 - x^2) Find d/dx: A'(x) = 8 - 6x^2 Critical Points:  0 = A'(x) = { -1.1547, 1.1547} ~~Then am I supposed to use this formula??? ----> 4x + 3y ?

-M
• Jul 22nd 2009, 04:55 PM
Prove It
Quote:

Originally Posted by mant1s
Hi everyone, I've been working on this optimization problem for awhile now. I guess I don't understand the steps. Here's my work.

The Problem
Code:

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 4 - x^2.  What are the dimensions of such a rectangle with the greatest possible area?
My work

Code:

Define variables... A = 2xy y = 4- x^2   setup problem:   A(x) = 2x ( 4 - x^2)   Find d/dx:   A'(x) = 8 - 6x^2   Critical Points:  0 = A'(x) = { -1.1547, 1.1547}   ~~Then am I supposed to use this formula??? ----> 4x + 3y ?

-M

You don't need to use the formula $\displaystyle 4x + 3y$ at all.

You are exactly right, the Area of the rectangle is given by

$\displaystyle A = 2x ( 4 - x^2)$

$\displaystyle = 8x - 2x^3$.

So to find the maximum

$\displaystyle \frac{dA}{dx} = 8 - 6x^2$

$\displaystyle 0 = 8 - 6x^2$

$\displaystyle 6x^2 = 8$

$\displaystyle x^2 = \frac{4}{3}$

$\displaystyle x = \pm \sqrt{\frac{4}{3}}$

$\displaystyle x = \pm \frac{2\sqrt{3}}{3}$.

The maximum area would be at the point $\displaystyle x = \frac{2\sqrt{3}}{3}$.

So put it into the formula for area.

$\displaystyle A = 8x - 2x^3$

$\displaystyle A = 8\left(\frac{2\sqrt{3}}{3}\right) - 2\left(\frac{2\sqrt{3}}{3}\right)^3$
• Jul 22nd 2009, 05:01 PM
mant1s
Thank you so much!
• Jul 22nd 2009, 05:07 PM
mant1s
I got 6.1584 for the MAX

How would you find the Length and Width of the rectangle after that?

-M
• Jul 22nd 2009, 06:37 PM
compliant
well, if you look at prove it's post, the value of x that gives you the maximum area is already there. when x is that value, what is y? whichever is the greater of the two values becomes the length, the other is the width.
• Jul 24th 2009, 05:47 AM
mant1s
So am I looking to do an perimeter calculation with my MAX (X)? Using the 4x + 3y formula. Also how do I get y, just algebra?

Thx!

-M
• Jul 24th 2009, 07:09 AM
compliant
well, you see, how you find y is already in the question itself.

y = 4 - x^2.

 From there you can find x through the formula for area, as HallsofIvy mentions below.[/edit]
• Jul 24th 2009, 07:27 AM
HallsofIvy
Notice that while the height of the rectangle is y, the base length is 2x, not just x. The original problem says nothing about the perimeter. I don't understand why you keep mentioning it.
• Jul 25th 2009, 05:26 PM
mant1s
hey sweet!

I got...

y= 2.66667
x= 2.3094