# Math Help - Minimizing values word problem- my head is spinning!

1. ## Minimizing values word problem- my head is spinning!

A fuel tank is being designed to contain 200 m^3 of gasoline; however, the maximum length tank that can be safely transported to clients is 16 m long. The design of the tank calls for a cylindrical part in the middle with hemispheres at the end. If the hemispheres cost $2/unit and the cylindrical wall costs$1/unit, find the radius and height of the cylinder part so that the cost of manufacturing the tank will be minimal. Give the answer correct to the nearest centimeter.

Here's what I have so far:

V= πr^2h + 4/3πr^3
200= πr^2h + 4/3πr^3
(200 - 4/3πr^3)/(πr^2)= h

SA= 4πr^2 + 2πrh
= 4πr^2 + 2πr[(200 - 4/3πr^3)/(πr^2)]
= 4πr^2 + (400πr - 8/3π^2r^4)/r^2

C= 8πr^2 + (400πr - 8/3π^2r^4)/(πr^2)
C'= 16πr - [16(r^3π + 75)]/3r^2 ??

I'm pretty lost and I'm sure I made several large errors because this is just not making sense. Any help is very much appreciated. Thanks!

2. Originally Posted by Goose
A fuel tank is being designed to contain 200 m^3 of gasoline; however, the maximum length tank that can be safely transported to clients is 16 m long. The design of the tank calls for a cylindrical part in the middle with hemispheres at the end. If the hemispheres cost $2/unit and the cylindrical wall costs$1/unit, find the radius and height of the cylinder part so that the cost of manufacturing the tank will be minimal. Give the answer correct to the nearest centimeter.

Here's what I have so far:

V= πr^2h + 4/3πr^3
200= πr^2h + 4/3πr^3
(200 - 4/3πr^3)/(πr^2)= h

SA= 4πr^2 + 2πrh
= 4πr^2 + 2πr[(200 - 4/3πr^3)/(πr^2)]
= 4πr^2 + (400πr - 8/3π^2r^4)/r^2

C= 8πr^2 + (400πr - 8/3π^2r^4)/(πr^2)
C'= 16πr - [16(r^3π + 75)]/3r^2 ??

I'm pretty lost and I'm sure I made several large errors because this is just not making sense. Any help is very much appreciated. Thanks!
The part in red is where you made the mistake!

You should get $S=4\pi r^2+\dfrac{400-\tfrac{8}{3}\pi r^3}{r}=4\pi r^2+400r^{-1}-\frac{8}{3}\pi r^2$.

So, $C=8\pi r^2+400r^{-1}-\frac{8}{3}\pi r^2$

Thus, $C^{\prime}=16\pi r-400r^{-2}-\frac{16}{3}\pi r$.

Now, set C' equal to zero to get $32\pi r^3=1200\implies r=\frac{1}{2}\sqrt[3]{\frac{300}{\pi}}$.

I leave it for you to find the value for h and then verify that $2r+h\leq 16$!!