what relative dimensions will maximize the volume

S

Smily

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May 27, 2006
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hi everyone
i have a problem, and i would be happy if someone will help me. :oops:
A container with a closed top is to be constructed in the shape of a right circular cylinder. If the surface area has a fixed value S, what relative dimensions will maximize the volume?
 
A container with a closed top is to be constructed in the shape of a right circular cylinder. If the surface area has a fixed value S, what relative dimensions will maximize the volume?
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You need to write down some things you know about volume and surface area, then combine the two equations:

S = 2(pi)r^2 + 2(pi)rh
h = (S - 2(pi)r^2)/((2pi)r)
V = (pi)r^2(h)
V = (pi)r^2((S - 2(pi)r^2)/(2pi)r)
V = S(pi)r^2/(2(pi)r) - 2(pi^2)r^4)/(2pi)r
V = (Sr)/2 - (pi)r^3)
dV/dr = S/2 – 3(pi)r^2

Please check my work. Can you take it from here?
 
so, dimension is dv/dr?
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DV/dr just tells you how volume is changing with respect to the radius changing.

How do you solve a max/min problem? Answer: find places where the function has zero slope. How? Take the derivative and set it to zero and solve.
 
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