Open Tank Cost Minimisation under a Height Restriction
A metal fabricator is designing an open-topped rectangular tank with a square base to hold 32 m³ of liquid. The base is made from reinforced sheet costing 3 currency units per square metre, while the four vertical sides use standard sheet costing 2 currency units per square metre. Because of a safety cage, the height of the tank must be at least 1 m and at most 3 m. The fabricator's apprentice sets up a cost function in terms of the base side x and differentiates it, obtaining a single stationary point, then claims this must be the minimum. The workshop supervisor is not convinced: the height restriction may mean the cheapest permitted tank is at an endpoint of the allowed interval. A quote is needed for the cheapest tank that satisfies the height rules. There is no lid, and welding costs are included in the sheet prices.
1. Express the cost C in terms of the base side x, and find the stationary point. State the corresponding height and check whether it satisfies the height restriction.
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Let the base side be x metres and the height be h metres. The volume is x^2 h = 32, so h = 32/x^2. The base area is x^2 and costs 3x^2. Each of the four sides has area xh, so the total side area is 4xh and its cost is 2(4xh) = 8xh. Hence C = 3x^2 + 8x(32/x^2) = 3x^2 + 256/x. Differentiate: dC/dx = 6x − 256/x^2. Setting this to zero gives 6x = 256/x^2, so 6x^3 = 256 and x^3 = 128/3 ≈ 42.667. Thus x ≈ 3.494 m. The corresponding height is h = 32/x^2 ≈ 32/12.21 ≈ 2.62 m, which lies between 1 m and 3 m, so the stationary point is permitted. Its cost is C ≈ 3(12.21) + 256/3.494 ≈ 36.63 + 73.27 = 109.9 currency units.
2. Examine the endpoints h = 1 and h = 3, compare costs, and decide whether the apprentice's stationary point is the global minimum. Explain the role of the restricted interval.
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At h = 1, x^2 = 32 so x = √32 ≈ 5.657 m. Then C = 3(32) + 256/5.657 ≈ 96 + 45.25 = 141.25. At h = 3, x^2 = 32/3 so x ≈ 3.266 m. Then C = 3(32/3) + 256/3.266 ≈ 32 + 78.38 = 110.38. The stationary value 109.9 is lower than both endpoint values, so it is the global minimum on the permitted interval 1 ≤ h ≤ 3. The apprentice is right in this case, but not for the stated reason: finding a single stationary point is not enough when the domain is restricted. The second derivative or the shape of the function can classify the stationary point, but the endpoints must still be checked because the cheapest permitted tank could occur at a boundary. The recommended dimensions are base side about 3.49 m and height about 2.62 m.