Thread: 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 ?

Thanks for your time

-M

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 ?

Thanks for your time

-M

You don't need to use the formula at all.

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



.


So to find the maximum











.


The maximum area would be at the point .


So put it into the formula for area.

Thank you so much!

I got 6.1584 for the MAX

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

-M

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.

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

well, you see, how you find y is already in the question itself.

y = 4 - x^2.

[edit] From there you can find x through the formula for area, as HallsofIvy mentions below.[/edit]

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.

hey sweet!

I got...

y= 2.66667
x= 2.3094