Calculus-derivative + word problem Can you check my procedure?

[stanton]

1. The problem statement, all variables and given/known data

A cylindrical can is to hold 4π cubic units of juice. The cost per square unit of constructing the metal top and the bottom is twice the cost of constructing the cardboard side. what are the dimensions of least expensive can?
calculus must be used logically to solve this problem and work should be shown.

2. Relevant equations


The total surface area of the can is: A = 2πrh + 2πr²
The volume of the can is: V = πr²h = 4π, then h = 4/r²


3. The attempt at a solution

connect volume and area equation, A and V.

so that; A = 2πr(4/r²) + 2πr²

We want the area of the top to be a minimum while
having the same volume. Therefore, we need to minimize
r and vary the height. We will take the derivative of the
total surface to accomplish this aim.

dA/dr = 8π(-1/r²) + 4πr = 0,

r = ³√2,

and since the cost of Area of top is twice the area of cardboard; that is,

2πr² = 2(2πrh), then h = r/2

The dimensions for minimum material cost is then:

r = ³√2, and h = 2 ³√2

I did untill this. And I will plug this h value and r value in area equation. But I am nut sure i am doing right. especially the part with [since the cost of Area of top is twice the area of cardboard;]
Pleas give me a hand.

[n!kofeyn]

What you minimized was the surface area, but what you need to minimize is the cost of constructing the can. You know this because the question asks what are the dimensions the give the least expensive can. So come up with a cost function, and then minimize it.

[LCKurtz]

Where you are going wrong is that you are trying to minimize the wrong thing. The problem does not ask you for minimum area. It wants minimum cost. So instead of writing a formula A = ... for area, you need to write a formula for cost:

C = ???

Hint: Assume the cardboard is $1 per square unit and the top and bottom are $2 per square unit. What would be the cost of the top and bottom? And the side? So you will have C = a formula with r and h in it. Then use your relation between h and r to get a formula for cost in terms of just r. Then go to work on dC/dr.

[n!kofeyn]

Where you are going wrong is that you are trying to minimize the wrong thing. The problem does not ask you for minimum area. It wants minimum cost. So instead of writing a formula V = ... for volume, you need to write a formula for cost.

Well that's what I just posted, and you mean "instead of writing a formula C = ... for cost", not volume. He still needs the volume equation to have the relation between h and r.

[LCKurtz]

Well that's what I just posted, and you mean "instead of writing a formula C = ... for cost", not volume. He still needs the volume equation to have the relation between h and r.

Yep, fixed typo.