Thread: optimization problem please help!!!

1. optimization problem please help!!!

A cylinder shaped can needs to be constructed to hold 300 cubic centimeters of soup. The material for the sides of the can costs 0.04 cents per square centimeter. The material for the top and bottom of the can need to be thicker, and costs 0.08 cents per square centimeter. Find the dimensions for the can that will minimize production cost.

Helpful information:
h : height of can, r : radius of can

Volume of a cylinder: V=πr2h

Area of the sides: A=2πrh

Area of the top/bottom: A=πr2

2. Originally Posted by eawolbert
A cylinder shaped can needs to be constructed to hold 300 cubic centimeters of soup. The material for the sides of the can costs 0.04 cents per square centimeter. The material for the top and bottom of the can need to be thicker, and costs 0.08 cents per square centimeter. Find the dimensions for the can that will minimize production cost.

Helpful information:
h : height of can, r : radius of can

Volume of a cylinder: V=πr2h

Area of the sides: A=2πrh

Area of the top/bottom: A=πr2
You have all the information needed to solve this question:

$V=300=\pi r^2 h~\implies~\boxed{h=\frac{300}{\pi r^2}}$

$A=2 \pi r^2 + 2 \pi r h$

Let C denote the costs:

$C= 0.08 \cdot 2 \pi r^2 + 0.04 \cdot 2 \pi r h$

Substitute h by term calculated above and you'll get a function dependent on r:

$C(r)= 0.08 \cdot 2 \pi r^2 + 0.04 \cdot 2 \pi r \cdot \frac{300}{\pi r^2}$

$C(r)= 0.16 \cdot \pi r^2 + \frac{24}{ r}$

Calculate the first derivative and solve C'(r) = 0 for r

I've got $r = \sqrt[3]{\frac{24}{0.32 \cdot \pi}}\approx 2.879$

Plug in this value into the equation of h.

3. thanks a lot really helped me.