# Optimization application prob

• May 2nd 2009, 08:40 AM
calvin_coolidge
Optimization application prob
A large storage tank is to be built in the shape of a cylinder. Assume the following facts about the cost:

Cost of the floor is $12/sq ft Cost of the curved side wall is 1.8 times the cost of the floor Cost of the roof is 2.6 times the cost of the floor a) If the total volume is fixed at 1.2 million cu ft, what is the cost of the cheapest tank to construct? b) Compute the ratio of height to diameter for the cheapest tank. I stink at these....any help would be appreciated. ~CC • May 2nd 2009, 11:48 AM running-gag Hi Let $V_0$ be the volume of the tank (1.2 million cu ft) Let h be the height of the tank and r its radius $V_0 = \pi r^2h \Rightarrow h = \frac{V_0}{\pi r^2}$ Let $C_f$ be the cost of the floor per unit surface ($12/sq ft)

The cost of the tank is the sum of :
- cost of the floor $\pi r^2C_f$
- cost of the curved side wall $2\pi rh \times 1.8C_f = 3.6\pi rhC_f$
- cost of the roof $\pi r^2 \times 2.6C_f = 2.6\pi r^2C_f$

$C = \pi r^2C_f + 3.6\pi rhC_f + 2.6\pi r^2C_f = 3.6\pi r^2C_f + 3.6\pi r\frac{V_0}{\pi r^2}C_f = 3.6C_f \left(\pi r^2 + \frac{V_0}{r}\right)$

Spoiler:

The extremum is obtained for
$\frac{dC}{dr}=0 \Rightarrow 2\pi r - \frac{V_0}{r^2}=0 \Rightarrow r = \left(\frac{V_0}{2\pi}\right)^{\frac13}$

$h = \frac{V_0}{\pi r^2} = \left(\frac{4V_0}{\pi}\right)^{\frac13}$
$\frac{h}{2r} = 1$
• May 7th 2009, 07:08 AM
calvin_coolidge
Thank you for the help on the cost function, however I'm not following what you did or doing with the height in the spoiler section. Is that the derivative of the height using the quotient rule?

Thanks

~CC
• May 7th 2009, 11:40 AM
running-gag
No
The height is given by the first formula I wrote
As soon as you have the radius, you can find the height