Question

A soup company wants to manufacture a can in the shape of a right circular cylinder that will hold 500 cm^3 of liquid. The material for the top and bottom costs 0.02 cent/cm^2, and the material for the sides costs 0.01 cent/cm^2.
a) Estimate the radius r and height h of the can that costs the least to manufacture. [suggestion: Express the cost C in terms of r.]

Answer 1

first find an expression the area of the can in terms of r and h hint: the area of the can is the area of two circles of radius r and one rectangle (think about 'unrolling' the side of the can to get an expression for this area)

Answer 2

So far I have \[500 = \pi r ^{2}h\] and SA=\[2 \pi r ^{2}+2 \pi rh\]

Answer 3

Good...then find an expression for the cost of the can: area of ends (the two circles) times .02 plus area of the side times .01

Answer 4

I don't understand how to do that part

Answer 5

That's the equation you're going to want to minimize (ie find the derivative and set it equal to zero)

Answer 6

area of the ends is 2*pi*r^2 so cost of material used on the ends is .02*2*pi*r^2

Answer 7

so Cost = .02*2*pi*r^2 +.01*2*pi*r*h

Answer 8

That part clear?

Answer 9

yes. now what do I do with those equations?

Answer 10

like I said, need to minimize cost.. so take the derivative and set equal to zero...

Answer 11

one little problem though.. it's an equation in two variables...

Answer 12

we'd like it to be in terms of only one variable (r)...

Answer 13

so... is there more info in the problem statement that you can use to get a second equation containing r and h?

Answer 14

there is :)

Answer 15

they tell you the volume of the can...

Answer 16

find that equation, then you'll see that you can use it to simplify the equation we found for Cost into an equation with only one variable (r)

Answer 17

so the 500=pi*r^2*h?

Answer 18

yes:)

Answer 19

solve for 'h' and sub.s into the cost equation we found...

Answer 20

500/(pr*r^2) = h right?

Answer 21

yes

Answer 22

okay now what do I do after I plug that into the cost equation?

Answer 23

you're in first year calc. right?

Answer 24

yes

Answer 25

ok, it's a minimization problem... you take the derivative of cost (with respect to radius) and set it equal to zero, remember? ... that tells you the equation for 'r' that gives you a max or min for cost...

Answer 26

we know here that when Cost' =0 we're going to have a minimum... so don't be concerned about determining whether Cost'=0 is a max...

Answer 27

I'm confused again. I don't think I'm supposed to be dealing with derivatives in this section and I forgot how to find the derivative of this function anyways....

Answer 28

Is the chapter named 'optimization' ?

Answer 29

No, it's just functions

Answer 30

the chapter is named 'functions' ??

Answer 31

yes and the section I'm working on is 'functions and the analysis of graphical information'

Answer 32

I think they probably expect you to take the derivative odf cost and set it equal to zero and then solve for r.... if not then there are other ways ... you could plot the cost function on your calculator or some software and trace until you are near the min...

Answer 33

yes then that's probably what they want you to do...

Answer 34

plot the cost function we found Cost(r) ... trace it to locate the minimum cost... find the r value at that minimum point

Answer 35

I'm just getting a line when I plug it into my calculator

Answer 36

use 'x' for 'r'

Answer 37

I did

Answer 38

it's not a line... it has a squared term and a 1/r term in it... make sure you entered it correctly... make sure the scale is zoomed in enough so that you can see features...

Answer 39

I plugged it in as \[(.02)2 \pi x ^{2} + (.01)2 \pi x(500/ \pi x ^{2})\]

Answer 40

sure.. you can simplify it a bit before you enter it... that might help

Answer 41

anyway the x window should be set to around 0...6

Answer 42

you'll see a definite minimum then with the y window set to 0..6 as well

Answer 43

I have something like this now