If the pen is x by y, then we have
xy=720
p = 2x+y = 2x+720/x
dp/dx = 2 - 720/x^2
x^2=360
x = 6√10
So, the pen is 6√10 by 12√10
As always, the area is max when the fence is divided equally among lengths and widths.
clearly the cost per foot of the fence does not affect the optimal dimensions.
If the fence costs $m/ft and the mesh costs $n/ft^2, then the cost is
c = m(2x+y)+720n
= 2mx + 720m/x + 720n
Clearly, neither the cost of the fence nor the cost of the mesh affects the dimensions
I suggest you explore a semicircular enclosure for lower cost.
You have been hired by a farmer to design a fenced-in rectangular enclosure for emus. The emus will require 720 square feet of area in which to roam, and the fence will cost 20 dollars per foot. The rectangular area will adjoin an existing wall, so a fence is only needed on three sides. You should minimize the cost of materials by choosing the dimensions carefully. What should the dimensions be, how much will it cost, and why? As you are writing the report for your client, you realize that the emus may try to bite through the fence. You may need a stronger fence that costs 30 dollars per foot instead. Also, although emus cannot fly, they can jump, so you may need to cover the top of the enclosure with a wire mesh to prevent their escape. The wire mesh costs 2 dollars per square foot. Would these upgrades affect the optimum dimensions? If so, how and why? If not, why not? How would these upgrades affect the cost? In case the client is unable or unwilling to pay for the rectangular enclosure that you are designing, you should also propose an alternative shape for the enclosure. (Some possibilities: more sides, fewer sides, different angles, a curved fence.) Make at least one proposal that costs less than the rectangular one. How much less will it cost, and why?
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