Open-Top Tank Design with a Roof Height Limit
A metal fabricator has a contract to make open-top tanks with a square base. Each tank must hold exactly 125 cubic metres. The base is cut from steel costing K2 per square metre; the four vertical sides use a cheaper steel costing K1 per square metre. Welding and labour are charged separately and do not depend on the dimensions. Let the base side be x metres and the height be h metres. The workshop roof limits any tank to a height of at most 4 metres. A junior draughtsman proposes a cube x = h = 5 m, saying it is the most economical shape. A supervisor is not convinced because the material prices for base and sides differ, and the roof limit may matter. The fabricator wants the dimensions that minimise material cost and a clear statement of how much the height limit costs if it is binding. Scrap and sheet thickness are ignored.
1. Express the height h in terms of x. Write the material cost C(x) and state the domain imposed by the roof height limit.
Read the worked answer
Volume is x^2h=125, so h=125/x^2. Base cost is 2x^2. The four sides have total area 4xh and cost at K1 per square metre, giving 4x(125/x^2)=500/x. Hence C(x)=2x^2+500/x, with x>0. The roof limit h≤4 gives 125/x^2≤4, so x^2≥31.25 and x≥sqrt31.25≈5.59 m. Thus the practical domain is x≥5.59 m.
2. Find the dimensions that minimise C(x) if the roof limit is ignored. Check whether this unconstrained optimum is allowed, and verify it is a minimum.
Read the worked answer
Differentiate: C'(x)=4x-500/x^2. Set C'=0: 4x=500/x^2, so 4x^3=500 and x^3=125, giving x=5. Then h=125/25=5. The cost is C(5)=2(25)+500/5=50+100=150. Also C''(x)=4+1000/x^3>0 for x>0, so x=5 is a minimum. But h=5 m exceeds the 4 m roof limit, so the unconstrained cube is not feasible in this workshop.
3. Find the minimum cost and dimensions when the height limit is enforced. How much extra material cost does the roof limit cause compared with the unconstrained optimum?
Read the worked answer
With h≤4, the smallest allowed x occurs when h=4: 125/x^2=4, so x^2=31.25 and x=5.590. For x≥5.590, C'(x)=4x-500/x^2. At x=5.590, C'≈22.36-500/31.25=22.36-16=6.36>0, and C' remains positive for larger x, so C increases from the boundary. Hence the constrained minimum is at x≈5.59 m and h=4 m. C≈2(31.25)+500/5.590=62.5+89.45=151.95. The extra cost is about 151.95-150=1.95, roughly K2. The roof limit forces a slightly wider, shorter tank and raises material cost by about K2.