# Thread: Max square/rec area under the curve

1. ## 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

2. 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 $4x + 3y$ at all.

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

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

$= 8x - 2x^3$.

So to find the maximum

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

$0 = 8 - 6x^2$

$6x^2 = 8$

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

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

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

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

So put it into the formula for area.

$A = 8x - 2x^3$

$A = 8\left(\frac{2\sqrt{3}}{3}\right) - 2\left(\frac{2\sqrt{3}}{3}\right)^3$

3. Thank you so much!

4. I got 6.1584 for the MAX

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

-M

5. 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.

6. 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

7. 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]

8. 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.

9. hey sweet!

I got...

y= 2.66667
x= 2.3094