PDA

View Full Version : Precalc Optimizing


kitteneater
Sep 9, 2008, 12:26 PM
I'm having problems with this problem...
"A rectangle has its base on the X-axis, and its upper two vertices lie on the parabola y=12-x^2. What is the largest area the rectangle can have? What are the dimensions?"

I've gotten this far...
Area= Length * Width
Length= 2X (cuz X can't go into the negatives, right?) * width
y=12-x^2
A=2x*(12-x^2)
Any time I try to graph it to see the Maximum, I see nothing.

Am I even on the right track?

ebaines
Sep 9, 2008, 01:15 PM
Yes, you're on the right rack. Are you graphing throughout the full domain for x from 0 to sqrt(12)? When you graph it, you're a function should look like the attached. Note there's a definite maximum.

kitteneater
Sep 9, 2008, 04:18 PM
Oh! Thank you so much!! I think I put it in wrong! So would the area be 32, and the dimensions be 2*32?

ebaines
Sep 10, 2008, 05:31 AM
Almost right - yes, the area is 32, and the max occurs when x = 2, but that makes the dimensions 4*8.