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Math Help - max area with reptiles

  1. #1
    Dec 2009

    Post max area with reptiles

    A herpetologist is designing a rectangular experimental iguana to fit in the 8ft by 20ft corner of a komodo dragon pen. If only 42 feet of fencing is available for the iguana pen, what dimensions should be used to obtain the largest area for the new fence.

    I got the dimensions 20ft by 15ft. I am not sure if that is right. How would all the steps look? I think I skipped one in the end.
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    Last edited by hotdogking; April 11th 2010 at 07:03 PM.
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  2. #2
    Super Member
    Mar 2010
    You want to maximize area = length * width with the condition that the amount of fencing = 2 * length + 2 * width - 8 - 20 = 42, so length + width = 35 and width = 35 - length. If you substitute that into the equation for area and set the derivative with respect to length equal to zero, you get the result length = width = 17.5. Unfortunately, that solution is not valid because the equation for the amount of fencing is invalid for this width.

    If the width is 20 or less, the amount of fencing is independent of w, so we should make w the maximum value of 20, which makes the length 15, and your solution is correct.

    If you're still having trouble, please post again in this thread.
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