Capsule tank material cost
A food-grade mixing capsule is made from a right circular cylinder of radius r cm and height h cm with two hemispherical end caps. Its total volume must be exactly 500 cm³. The cylindrical wall costs $0.02 per cm², the hemispherical caps cost $0.03 per cm², and there is a fixed seam charge of $8 that does not depend on r or h. The workshop can manufacture any capsule with 2 ≤ r ≤ 5 and h ≥ 0. A junior designer proposes r = 2 because it reduces the expensive curved cap area; another proposes r = 5 because the cylinder becomes short. The production manager asks for the radius that minimizes variable material cost and whether the fixed seam charge changes that choice. The manager also needs the exact feasible interval for r imposed by h ≥ 0 before authorizing a prototype.
1. Express h in terms of r, and write the variable material cost C(r). State the exact feasible interval for r.
Read the worked answer
Volume: πr²h + (4/3)πr³ = 500, so h = 500/(πr²) - 4r/3. For h ≥ 0, 500/(πr²) ≥ 4r/3, hence r³ ≤ 375/π and r ≤ (375/π)^(1/3) ≈ 4.92. Together with 2 ≤ r ≤ 5, the feasible interval is 2 ≤ r ≤ (375/π)^(1/3) ≈ 4.92.
Variable cost: C = 0.02(2πrh) + 0.03(4πr²) = 0.04πrh + 0.12πr². Substituting h gives C(r) = 0.04πr(500/(πr²) - 4r/3) + 0.12πr² = 20/r + πr²/15. The $8 seam charge adds a constant and does not alter the minimizer.
2. Find the radius that minimizes variable material cost on the feasible interval. Check the endpoints and explain the role of the fixed charge.
Read the worked answer
C'(r) = -20/r² + 2πr/15. Setting C'(r)=0 gives 2πr/15 = 20/r², so r³ = 150/π and r = (150/π)^(1/3) ≈ 3.63 cm, which lies in the feasible interval. C''(r)=40/r³ + 2π/15 > 0, so this critical point is a minimum.
Endpoint checks: C(2)=10+4π/15≈10.84; C(4.92)≈9.13; at r≈3.63, C≈8.27. Therefore the global minimum on the feasible interval is at r=(150/π)^(1/3)≈3.63 cm. The fixed $8 changes total cost but not the optimal radius.