Course:MATH102/Question Challenge/2003 December Q7
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Question
A quonset hut consists of half a cylinder lying on its side. If the perimeter of the rectangular base of the hut is 60 meters, what is the maximum possible volume? (Note that the volume of a cylinder of radius r and height h is .)
Hints
Hint |
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The equation which describes the perimeter of the rectangular base is:
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Solutions
Solution |
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= volume of a cylinder with radius and height = the volume of a quonset hut.
Before we can maximize the volume of the hut, we must first eliminate a variable by modifying the constraint:
After substituting this value in for h we can find the maximum volume of the hut by taking the first derivative:
To prove that this is a maximum volume and not a minimum we substitute into the second derivative of the hut's volume:
Since the second derivative is negative, the critical point is a local maximum. The next step is to substitute into the constraint to solve for :
To find what the maximum volume of the hut is, we substitute in these values of and :
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