Question

I need help figuring out this one... From square one, I just don't know what the heck it's talking about, but beyond that, how does one solve this analytically? The problem:

"A rectangle has its base on the x-axis and its upper two vertices on the parabola 'y = 12 - x^2'. What is the largest area the rectangle can have?"

Any and all help is greatly appreciated!! :)

Answer 1

Hey @wolfe8, you up for a little Calculus? ;)

Answer 2

Can I give you a link that might help? http://answers.yahoo.com/question/index?qid=20090404100038AATDsJu

Answer 3

. . . god, I do love yahoo answers, but I'm using a computer issued by my school which unfortunately has internet blocks to suspect blogging/ social sites. Sorry! But no... I can 'to use that :/

Answer 4

Can you copy and paste text from the we page into a reply?

Answer 5

Sure. And sorry in advance for not going through it yourself. I would if my brain would cooperate.

The parabola is symmetric around x=0. Therefore, if (x,12-x²) is one of the vertices of the rectangle, so is (-x, 12-x²). Therefore, the width of the rectangle is x - (-x) = 2x and its height is 12-x².

You want to maximize the area A = 2x(12-x²) so take a derivative and set it equal to 0: 24-6x² = 0 or x²=4 leading to the solution of x=2. Thus, the width is 4, the height is 8, and the area is 32.

Answer 6

myself*

Answer 7

It's okay :) I understood ;)

Answer 8

How, analytically, do we know that the graph is symmetric around the Y-axis?

Answer 9

Perhaps by using the vertex equation?

Answer 10

And what're the base equations used? What did they take the derivative OF? I think I get the concept they're shooting at here, but what's the work? Can I seee some algebra? That's something I can fully conceptualize and understand. I know it can be tedious to type out, but it would really help me out! :)

Answer 11

The height of the rectangle is can be found by taking x=0.
The width of the rectangle can be found by taking y =0.


So Area = 12 * 2*sqrt(12)