Question invovling cyclinderical volume and such...

• Oct 25th 2005, 04:57 PM
Zell_49
Question invovling cyclinderical volume and such...
Ok. I am stumped in my calculus class and need help with one of the problems. I am supposed to find the dimensions (r and h) of the cylinder that will minimize the cost of materials, knowing that V is constant, if the cost of the top and bottom are .15 per square inch and the side is .07 per square inch. (Note that my answer will be in terms of V.). help?
• Oct 30th 2005, 10:48 PM
ticbol
"Ok. I am stumped in my calculus class and need help with one of the problems. I am supposed to find the dimensions (r and h) of the cylinder that will minimize the cost of materials, knowing that V is constant, if the cost of the top and bottom are .15 per square inch and the side is .07 per square inch. (Note that my answer will be in terms of V.). help?"

Umm, I didn't see this question before.

One way to solve this is to express the Cost C in terms of a related variable, say radius r, and then set dC/dr = 0

Cost C = 2(pi r^2)*(0.15) +(2pi r h)(0.07)
C = (0.3pi)(r^2) +(0.14pi)(r)(h) -----------(i)
We try to eliminate the h so that C will be in terms of r only.

Volume of cylinder, V = (pi r^2)(h)
So, h = V / (pi r^2) ---------------------(ii)
Substitute that into (i),
C = (0.3pi)(r^2) +(0.14pi)(r)(V / pi r^2)
C = (0.3pi)(r^2) +(0.14V)/r ------------(iii)
There. There is only one variable for C. Remember that V is constant as mentioned.

Differentiate both sides of (iii) with respect to r,
dC/dr = (0.6pi)(r) +(0.14V)(-1/ r^2)
dC/dr = (0.6pi)(r) -(0.14V)/(r^2)
Set dC/dr to zero,
0 = (0.6pi)(r) -(0.14V)/(r^2)
(0.14V)/(r^2) = (0.6pi)(r)
Clear the fraction, multiply both sides by r^2,
0.14V = (0.6pi)(r^3)
r^3 = (0.14V)/(0.6pi)
r = cuberoot(0.14V / 0.6pi)