Minimization Problem.

D

djdownfawl

New member
Joined
Sep 28, 2010
Messages
37
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.
 
djdownfawl said:
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!
Click to expand...

The volume is given by \(\displaystyle {\pi}r^{2}h=75\).........[1]

Since cost is proportional to surface area, we have:

The cost is \(\displaystyle S=k(\overbrace{2{\pi}rh}^{\text{sidewall}}+\underbrace{2{\pi}r^{2}}_{\text{top and bottom}})\)..........[2]

Surface area is what must be minimized in order to arrive at the cheapest cost. Solve [1] for h, or even r^2, and sub into [2].

Then, S will be in terms of one variable and you can minimize.

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.
Click to expand...

Let k=cost of sidewall of can.

Same as above except the total cost would be \(\displaystyle S=k\cdot 2{\pi}rh+3k\cdot 2{\pi}r^{2}\)
 
oh wait sorry i saw k is the cost.

but then do i take the derivative?
how..? i am confused how you take the der of pi r and k?
 
You don't take the derivative of Pi and k. They are constants. Differentiate S with respect to r or h(depending on which you solve for) after you make the sub.
 
So i came up with

S = -k 4 pi r (75/pi r^3) + k 4 pi r

Then

S = -k 4 (75/r^2) + k 4 pi r

Then

S= -300k/r^2 + k 4 pi r

Is that correct?

Next?
 
The derivative of the first problem, after subbing in \(\displaystyle h=\frac{75}{{\pi}r^{2}}\) from [1], is

\(\displaystyle S'(r)=4k{\pi}r-\frac{150k}{r^{2}}\)

Set to 0 and solve for r. h follows from above.

A tip: The minimum is achieved when the diameter and the height are the same.
 
ok so then i got rsqr = diameter = 4.56 and h = 4.59

then i need help finding the der of the problem 2.
 
You must log in or register to reply here.
Top