Constrained Material Cost for an Open-Top Tank
An open-top holding tank must have a square base of side x metres and height h metres, with volume 32 m³. The base is made of 40 USD/m² plate and the four sides of 25 USD/m² sheet. The workshop ceiling limits h to 2.0 m, and the base side cannot exceed 5.0 m because of the available floor bay. The fabricator first ignores the height limit, minimizes the material cost using calculus, and obtains a stationarity condition; the resulting height is then checked against the ceiling. If it violates the limit, the constrained optimum must be found on the boundary h = 2.0 m and compared with the feasible endpoint x = 5.0 m. A supervisor argues that because the unconstrained cost curve has a single stationary point, it must also be the constrained minimum. The shop needs the dimensions, the minimum cost, and a statement of whether the ceiling changes the answer.
1. Formulate the cost as a function of x, find the unconstrained stationary point, and classify it.
Read the worked answer
The volume constraint gives x²h = 32, so h = 32/x². The base area is x² and the four side areas total 4xh. Therefore
C(x) = 40x² + 25(4x)(32/x²) = 40x² + 3200/x, with x > 0.
Differentiate: C′(x) = 80x − 3200/x². Setting C′ = 0 gives 80x³ = 3200, so x³ = 40 and x = 40^(1/3) ≈ 3.42 m. Then h = 32/(3.42)² ≈ 2.74 m, and C ≈ 40(11.70) + 3200/3.42 ≈ 468 + 936 ≈ 1404 USD. Since C″(x) = 80 + 6400/x³ > 0 for x > 0, this stationary point is a minimum of the unconstrained cost curve.
2. Apply the ceiling constraint h ≤ 2.0 m and find the feasible minimum cost.
Read the worked answer
The unconstrained height 2.74 m exceeds the 2.0 m ceiling, so that design is not feasible. From h = 32/x² ≤ 2.0, we need x² ≥ 16, hence x ≥ 4.0 m. The feasible interval is 4.0 ≤ x ≤ 5.0 m.
On this interval, C′(x) = 80x − 3200/x². At x = 4.0, C′ = 320 − 200 = 120 > 0, and C′ increases further for larger x, so C is increasing throughout [4.0, 5.0]. The minimum is therefore at x = 4.0 m, h = 2.0 m:
C = 40(16) + 3200/4 = 640 + 800 = 1440 USD.
At x = 5.0 m, C = 1000 + 640 = 1640 USD. The ceiling changes the answer: the constrained minimum is 1440 USD, not the unconstrained 1404 USD, and the supervisor’s argument fails because the unconstrained stationary point is outside the feasible region.
3. If the ceiling were raised to 3.0 m, what would the optimal dimensions and cost be?
Read the worked answer
With h ≤ 3.0 m, the unconstrained height 2.74 m is now admissible. The base side remains x ≈ 3.42 m, which is below the 5.0 m floor-bay limit, so no boundary is active. The minimum cost is the unconstrained value:
C ≈ 40(3.42)² + 3200/3.42 ≈ 468 + 936 ≈ 1404 USD, with x ≈ 3.42 m and h ≈ 2.74 m.
The reduction from 1440 USD to about 1404 USD is modest, about 36 USD, because the ceiling constraint was only slightly binding. The exact stationary value satisfies x³ = 40; substituting gives C = 40x² + 3200/x = 40x² + 80x² = 120x², since 3200/x = 80x² at the stationary point. Thus C = 120(40)^(2/3) ≈ 1403.5 USD.