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Calculus/Calculus probem

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Expert: - 5/30/2009


Question

Picture of Question
A 5,000 mē rectangular area of a field is to be enclosed by a fence, with a moveable inner fence built across the narrow part of the field, as shown.The perimeter fence costs $10/m and the inner fence costs $4/m. Determine the dimensions of the field to minimize the cost.


Answer
Hi George,
Let the fence have dimensions of length l and width w, then the length of the movable inner fence is also w.
The area of the rectangular area is 5000 mē and so
lw = 5000
w = 5000/l
The total length of the perimeter fence is 2(l + w) and the total cost at $10/m is $10.2(l + w) = 20(l + w)
The cost of the movable fence would be $4.w = 4w
The total cost of fencing is therefore
C = 20(l + w) + 4w
 = 20l + 24w
 = 20l + 24(5000/l)
 = 20l + 120000/l
The minimum cost occurs when dC/dl = 0
dC/dl = 20 - 120000/lē
equating to zero
20 - 120000/lē = 0
20 = 120000/lē
lē = 6000
l = 77.46m
and
w = 5000/77.46
 = 64.55m

Regards
    Questioner's Rating
    Rating(1-10)Knowledgeability = 10Clarity of Response = 10Politeness = 10
    CommentWow that was a great answer! Very clear and concise. Thank you very much!


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Ahmed Salami

Expertise

I can provide good answers to questions dealing in almost all of mathematics especially from A`Level downwards. I believe i would be very helpful in calculus and can as well help a good deal in Physics with most emphasis directed towards mechanics.

Experience

An engineering graduate. I have been doing maths and physics all my life.

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