Applied Optimization Help  TOPIC_SOLVED

Limits, differentiation, related rates, integration, trig integrals, etc.

Applied Optimization Help

Postby wilkes2 on Tue Aug 13, 2013 5:20 am

Hello,

First time and really need the help. I am totally stumped on this question and I have been beating my head against the table all night. Basically I need to optimize a fence within a fenced area. There needs to be 6 feet between the two and I have 1000 feet of fence to build both fences. I have tried assigning variables in terms of l and w along with l-12 and w-12 but I always get stuck. Any help is greatly appreciated. Thank you.
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Postby stapel_eliz on Tue Aug 13, 2013 3:04 pm

wilkes2 wrote:I need to optimize a fence within a fenced area. There needs to be 6 feet between the two and I have 1000 feet of fence to build both fences. I have tried assigning variables in terms of l and w along with l-12 and w-12 but I always get stuck.

Please reply with the full and exact statement of the exercise, the complete instructions, and a clear listing of your efforts (beginning with the variable assignment and going all the way to where you "always get stuck"). Thank you! :wink:
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Re: Applied Optimization Help

Postby wilkes2 on Tue Aug 13, 2013 8:27 pm

The problem: You have 1000 feet of fencing to build an fence surrounded by another fence with 6 feet of space between. What is the maximum area of an enclosure you can build?

Basically it is a square within a square. I have tried to assign variables with the inner polygon as L and W and the outside variables as L+6 and W+6. However, everytime I try to solve for either L or W and then plug that into the area formula of A=(L+6)(W+6). I have tried even A=(L+6)(W+6)-LW. Then I try to find the critical points and absolute max and I end up with some odd numbers that 1) Do not check to calculate to a total perimeter of 1000 and 2) nor even make sense given the length of material. Logically speaking, the ideal optimization would be a square within a square. If you split the footage in two and expand one square by 6 each side and subtract 6 from the inner square you would get an outside perimeter of 524 feet and an inner perimeter of 476, which gives you the 6 feet between fences. Hence, the outer fence would be 131x131 and the inner would be 119x119. But given this is calculus, I doubt the solution is that simple given this is an optimization question.
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Re: Applied Optimization Help  TOPIC_SOLVED

Postby nona.m.nona on Wed Aug 14, 2013 11:16 am

wilkes2 wrote:The problem: You have 1000 feet of fencing to build an fence surrounded by another fence with 6 feet of space between. What is the maximum area of an enclosure you can build?

Basically it is a square within a square. I have tried to assign variables with the inner polygon as L and W and the outside variables as L+6 and W+6.

Since the area is specified (though unstated in the above) to be a square, then the length is equal to the width; thus, only one variable is required. Since a lower-case "L" is often mistaken for the digit "1", I will use "w" for "width" instead, with the understanding that w = L.

wilkes2 wrote:However, everytime I try to solve for either L or W...

In the above, you have provided no equation. Thus, what are you "solving"?

As you know the area to be square, you therefore know the inner and outer perimeters to be 4w and 4(w + 6), respectively. Sum these, and set equal to the given total length of fencing. This is an equation in one variable. Solve for the value of the variable. Since you were given that the area is a square, you will be done at this point.
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