The can that cannot be made small enough
A beverage company wants a cylindrical can with volume 500 cm^3. The side material costs €0.02 per cm^2 and the top and bottom material costs €0.05 per cm^2. The production line can form cans with radius r between 4 cm and 10 cm inclusive; a smaller radius is not possible with the current machinery. A junior engineer writes the cost as a function of r, differentiates, sets the derivative to zero and reports an optimal radius of about 3.17 cm with height about 15.8 cm. The production manager points out that this radius cannot be manufactured. The engineer replies that the calculus solution is the true minimum and that the constraint is only a technical detail. The company needs the minimum material cost among cans that can actually be produced, and wants to know how much the constraint costs compared with the unconstrained optimum.
1. Express the material cost C as a function of the radius r, using the volume constraint to eliminate the height.
Read the worked answer
Let r be the radius and h the height in cm. Volume: πr^2h=500, so h=500/(πr^2). The side area is 2πrh and the top-plus-bottom area is 2πr^2. Thus
C(r)=0.02(2πrh)+0.05(2πr^2) =0.04πr h+0.10πr^2.
Substitute h:
C(r)=0.04πr·500/(πr^2)+0.10πr^2 = 20/r + 0.10πr^2, for r>0.
The production constraint is 4<=r<=10.
2. Find the unconstrained critical point and classify it. Explain what the junior engineer has actually shown.
Read the worked answer
Differentiate:
C'(r)=-20/r^2+0.20πr.
Set C'(r)=0: 0.20πr=20/r^2, so r^3=100/π, r=(100/π)^(1/3)≈3.17 cm. Also C''(r)=40/r^3+0.20π>0 for r>0, so this is a local and global minimum of the unconstrained function on r>0. The corresponding height is h=500/(πr^2)≈15.8 cm, and C≈€9.47. The engineer's calculus is correct for the unconstrained problem, but it is not yet a feasible answer.
3. Find the minimum cost among cans that can actually be produced, and compare it with the unconstrained optimum. What judgement does the derivative alone miss?
Read the worked answer
Because r must be at least 4, the unconstrained critical point r≈3.17 is outside the feasible interval. On [4,10], C'(r)=-20/r^2+0.20πr. At r=4, C'(4)=-20/16+0.8π=-1.25+2.513≈1.26>0. For r>=4>3.17, C'(r)>0, so C is increasing on the whole feasible interval. Therefore the minimum feasible cost occurs at the smallest allowed radius, r=4 cm. Then h=500/(π·16)=125/(4π)≈9.95 cm. The cost is C(4)=20/4+0.10π·16=5+1.6π≈€10.03. The constraint costs about €10.03-€9.47=€0.56 per can compared with the infeasible unconstrained optimum. The derivative alone would give an answer that cannot be manufactured; the boundary must be checked.