Mathematics

Problem

A cylindrical container with a closed bottom and a removable lid must hold exactly

1000\,\text{cm}^3

of liquid. The side (lateral surface) of the cylinder is to be made from a material that costs \

0.02 per square centimeter, while the top and bottom are to be made from a stronger material that costs

0.05 per square centimeter. Let the radius of the base be

r

centimeters and the height be

h

centimeters. (a) Write formulas for the total material cost

C

and for the volume

V

in terms of

r

and

h

. (b) Use the volume requirement to express

C

as a function of a single variable. (c) Determine the critical point(s) of the cost function and find the corresponding dimension values

r

and

h

(to two decimal places). (d) Justify that the critical point found in part (c) yields the minimum possible material cost for the container. (e) Calculate the minimum material cost, in dollars, required to build the container at the dimensions you found.

Intuitive explanation

This is an optimization problem where you'll use the volume constraint to eliminate one variable, then minimize cost as a function of the other. The key insight is that the total cost depends on three surfaces with different price tags—the lateral surface ( $2\pi rh$ ) is cheap, while the top and bottom circles ( $2\pi r^2$ ) are expensive—so you want to balance making the cylinder tall and thin versus short and wide. Set up cost as a function of $r$ alone using $V = \pi r^2 h = 1000$ , then find where $\frac{dC}{dr} = 0$ .

Intuitive explanation

Browse the problem library