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 $V=\pi r^{2}h$.)

A quonset hut.

Hints

[show]Hint
The equation which describes the perimeter of the rectangular base is: $60=4r+2h$

Solutions

[show]Solution

$V=\pi r^{2}h$ = volume of a cylinder with radius $r$ and height $h$ $V(hut)={\frac {\pi }{2}}r^{2}h$ = the volume of a quonset hut. $60=4r+2h$ Before we can maximize the volume of the hut, we must first eliminate a variable by modifying the constraint: $60=4r+2h$ $2h=60-4r$ $h=30-2r$ After substituting this value in for h we can find the maximum volume of the hut by taking the first derivative: $V(hut)={\frac {\pi }{2}}r^{2}(30-2r)$ $V(hut)=15\pi r^{2}-\pi r^{3}$ $V'(hut)=30\pi r-3\pi r^{2}$ $30\pi r=3\pi r^{2}$ $r=10$ To prove that this is a maximum volume and not a minimum we substitute $r=10$ into the second derivative of the hut's volume: $V'(hut)=30\pi r-3\pi r^{2}$ $V''(hut)=30\pi -6\pi r$ $V''(hut)=30\pi -6\pi (10)=-30\pi <0$ Since the second derivative is negative, the critical point is a local maximum. The next step is to substitute $r=10$ into the constraint to solve for $h$: $60=4(10)+2h$ $60-40=2h$ $2h=20$ $h=10$ To find what the maximum volume of the hut is, we substitute in these values of $r$ and $h$: $V(hut)={\frac {\pi }{2}}r^{2}h$ $V(hut)={\frac {\pi }{2}}\cdot (10)^{2}\cdot (10)$ $V(hut)={\frac {\pi }{2}}\cdot 1000$ $V(hut)={\frac {(1000\pi )}{2}}$ $V(hut)=500\pi m^{3}$ $V(hut)=1570.80m^{3}$