How would you solve this? I'm confused. Please help :(

Question

If you look more closely at a real can, you will see that the top and bottom are formed by cutting disks with radius larger than r and folding them down. Taking this into account would increase the ratio more towards what we see in real cans. But more significantly, we ought to take into account the cost of manufacture of the can. Let’s assume most of the expense is incurred in joining the sides to the rims of the cans. If we cut the discs from hexagons (as above), then the total cost is proportional to 4√ 3r^2 + 2πrh + k(4πr + h) where k is the reciprocal of the length that can be joined for the cost of one unit area of metal. Show that this expression is minimized when cube root√V/k = cube root√πh/r · 2π − (h/r) / (πh/r) −4√3

Answer 1

Hi Abigail!

This is how I would explain this problem to you in tutoring. However, during tutoring, I would train you to be able to get this answer yourself, which is what really matters!

Before my explanation, my suggestion to you is to get comfortable with that equation, C = 4*√(3)*r^2 + 2πrh + k(4πr + h)

Do you understand that equation? Do you see where each part of the expression comes from? Just getting that far may get you a long way towards the final answer.

Now for the answer. This problem is solved by finding the minimum cost. The way that this minimization request is related to calculus is that, for piecewise differentiable functions, minima occur at critical points. Thus, it is enough to take the derivative of the cost function with respect to a relevant variable and find at what point the minimum occurs.

Often, with optimization problems, it is necessary to bring in another equation to make the variable we seek to optimize be a function of just one variable. In this case, C is a function of r and h, and the relationship between r and h is not clear. The key is, we must bring the volume into the problem. V = πr^2*h, and V is constant since we are talking about a can of fixed volume. So let's solve the volume equation for h and substitute into the cost function:

h = V/(πr^2)

C = 4√(3)*r^2 + 2πrV/(πr^2) + k(4πr + V/(πr^2))

C = 4√(3)*r^2 + 2V/(r) + k(4πr + V/(πr^2))

Now we can differentiate the function of C with respect to r and set that derivative equal to 0.

dC/dr = 8*√(3)*r - 2V/r^2 +k*4π - k*2V/(πr^3) = 0