1. ## Minimization Problem.

Problem A
What is the cheapest soup can that will hold 75in3 (cubic inches) of soup? Assume the can is cylindrical in shape and that the total cost of a can is proportional to the amount of material (tin) measured as area (in other words, don’t worry about the thickness of the tin). Be sure to provide both the diameter (or radius) and height of the can and sketch what this cheapest can looks like!

Problem B
Repeat the above problem but for the case where, for whatever reason, the cost of manufacturing the two lids of the can is three times higher than manufacturing the sidewall. Give your answer in the same form as above, including the sketch.

2. The volume of a cylinder is given by:

$\displaystyle V = \pi r^2h$

The surface area is given by:

$\displaystyle A = 2\pi r^2 + 2\pi rh$

Find the derivative of A, using the volume given to express either the radius in terms of height or vice versa.

Then, substitute back the radius or height in the volume formula to get the other dimension.

Can you do this part first?

3. You mean 4pir+pirh?

4. What do you mean? For the surface area of the cylinder?

No, I meant what I meant.

The area of a circile is $\displaystyle \pi r^2$

In a cylinder, there are 2 circles for two cross sections, hence $\displaystyle 2 \pi r^2$

Then, the curved surface, if opened has the length h and the width equal to the circumference of a circle, which gives: $\displaystyle 2\pi r h$

6. What do you not understand?

7. You want me to find the der of A = 2 pi r^2 + 2 pi r h?

8. You want to have a volume of 75 in^3 in a can having the lowest possible surface area so that the cost of the can is minimised yes?

Then, you need to know how does this area varies as r and h vary, while keeping the volume constant.

Hence, you are looking for the minimum point of the graph showing how A varies with either h or r.

9. Can you please show me the steps.. i really am bad at match i need help need to submit this...

10. As I told you in my first post in this thread, the volume, V, of a cylinder is given by

$\displaystyle V = \pi r^2h$

and the area of the same cylinder is given by:

$\displaystyle A = 2\pi r^2 + 2\pi rh$

We know that the volume is 75 in^3. We are looking for the minimum A while r or h vary. Let's look for the value of r, the radius of the circular cross section of the cylindrical can.

So, from the information we got, we have:

$\displaystyle V = 75\ in^3$

$\displaystyle \pi r^2h = 75$

$\displaystyle h = \dfrac{75}{\pi r^2}$

In the Area equation, we get:

$\displaystyle A = 2\pi r^2 + 2\pi rh$

Substitute h by what we just got;

$\displaystyle A = 2\pi r^2 + 2\pi r\left(\dfrac{75}{\pi r^2}\right)$

Can you simplify this and find it's derivative?