Challenging cylindrical problem with a derivate solution (no calculator allowed)

Question

A closed cylindrical can with radius r centimetres and height h centimetres has a volume of 20π cm³. (Express the function as h in terms of r)

The material for the base and top of the can costs 10 cents per cm² and the material for the curved side costs 8 cents per cm³ . The total cost of the material, in cents, is C. Show that C =20πr² + 320π / r

Also, given that there is a minimum value for C, find this minimum value in terms of π .

Answer 1

Since material costs are per unit area, we need to find the areas of the various parts of the can and multiply them by the cost per area:

Cost (C) = (area of top and bottom)·(10 cents) + (area of sides)·(8 cents)

C = (πr² + πr²)(10) + (2πrh)(8)

C = 20π r² + 16πhr

C is now in terms of two variables, r and h. We want to eliminate the h so that C is in terms of one variable (r) only. We use the given constraint on the volume of the can:

Volume = area of bottom x height

20π = πr²h

20π/πr² = h

20/r² = h

Now substitute 20/r² in place of h in the cost equation:

C = 20π r² + 16πr(20/r²) = 20π r² + 320π/r

C is now expressed in terms of a single variable, r. This allows us to find the minimum cost by taking the first derivative of C wrt r, setting it to zero, then solving for r. Once you have the value of r, plug it into the cost equation to get the minimum cost.