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    kitteneater's Avatar
    kitteneater Posts: 4, Reputation: 1
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    #1

    Sep 9, 2008, 12:26 PM
    Precalc Optimizing
    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's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    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.
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    kitteneater's Avatar
    kitteneater Posts: 4, Reputation: 1
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    #3

    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's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #4

    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.
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