Minimum-cost tank geometry under a plate-width limit
A fabrication unit has won an order for four identical open-top rectangular steel tanks, each of internal volume exactly 4.00 m3. The owner's costing sheet is: base plate at Rs 450 per m2, side plates at Rs 300 per m2, and four vertical corner welds at Rs 120 per running metre each, that is Rs 480 per metre of tank height. No lid is needed and the plate thickness is constant, so cost depends only on geometry.
The foreman has already sketched a tank 2.00 m long, 1.00 m wide and 2.00 m deep, arguing that a tall tank uses less plate. The owner is not convinced and wants the cheapest practical size for the whole order.
Practical limit: the shop's plate bender accepts a maximum developed width of 1.80 m, so neither plan dimension may exceed 1.80 m. The crane has 3 m of hook height, so height is unrestricted. Each tank stands on a flat floor with a rectangular base of sides x and y and height h, all positive.
1. Ignoring the 1.80 m bender limit, show that for a fixed height the cheapest base is square, write the cost per tank as a function of h alone, and locate the unconstrained optimum.
Read the worked answer
Volume constraint: xyh = 4, so xy = 4/h.
Cost C = 450xy + 300[2h(x + y)] + 480h = 450xy + 600h(x + y) + 480h.
For a fixed h the first term is fixed at 1800/h and the coefficient 600h is positive, so C is minimised when x + y is minimised subject to xy = 4/h. By AM-GM, x + y >= 2 sqrt(xy), with equality at x = y. So the cheapest base is square, x = y = 2/sqrt(h).
Substituting: C(h) = 1800/h + 2400 sqrt(h) + 480h.
C'(h) = -1800/h^2 + 1200/sqrt(h) + 480. Setting C' = 0 and putting u = sqrt(h) gives 4u^4 + 10u^3 - 15 = 0, whose positive root is u = 1.021, so h = 1.04 m.
Then x = y = 2/1.021 = 1.96 m and C = 1727 + 2450 + 500 = Rs 4,677 per tank.
The unconstrained optimum therefore needs a 1.96 m plan dimension, which the bender cannot form.
2. Apply the 1.80 m bender limit and find the true optimum. Compare it with the foreman's 2.00 x 1.00 x 2.00 m tank over the full order of four tanks.
Read the worked answer
The unconstrained optimum is infeasible because 1.96 m exceeds 1.80 m. Cost falls as x and y grow towards that point, so the true optimum lies on the boundary.
Try one dimension at the limit, x = 1.80 and y free: C(h) = 450(4/h) + 600h(1.80 + 4/(1.80h)) + 480h = 1800/h + 1560h + 1333. C'(h) = -1800/h^2 + 1560 = 0 gives h = 1.074 m, and then y = 4/(1.80 x 1.074) = 2.07 m, which again breaks the limit.
So both plan dimensions bind: x = y = 1.80 m, h = 4/1.80^2 = 1.235 m. C = 450(3.24) + 300 x 2 x 1.235 x 3.60 + 480 x 1.235 = 1458 + 2667 + 593 = Rs 4,717.
Foreman's tank: C = 450(2.00) + 300 x 2 x 2.00 x 3.00 + 480 x 2.00 = 900 + 3600 + 960 = Rs 5,460.
Order cost: 4 x 4717 = Rs 18,868 against 4 x 5460 = Rs 21,840, a saving of Rs 2,972 (13.6%). The tall tank is the expensive one, because the weld cost and the two large sides outweigh the saving in base plate.
3. The owner asks whether the last few rupees are worth chasing. Quantify how flat the cost curve is near the optimum and state what you would recommend building.
Read the worked answer
Along the feasible boundary the cost is very insensitive. Take nearby feasible shapes of the same 4 m3 volume:
- h = 1.30 m, x = y = sqrt(4/1.30) = 1.754 m: C = 450(3.077) + 300 x 2 x 1.30 x 3.508 + 480(1.30) = 1385 + 2737 + 624 = Rs 4,746.
- h = 1.10 m needs x = y = 1.907 m, which breaks the 1.80 m limit, so it is not available.
A 5% change in height, with the plan size adjusted to hold volume, moves cost by only about Rs 30, under 0.7%. The cost surface is nearly flat in h near the constrained optimum, so precision in h buys nothing.
Recommendation: build 1.80 x 1.80 x 1.24 m, and round the height to 1.24 m rather than 1.235 m. Do not redesign to chase Rs 20. Effort is far better spent on plate nesting and on shortening the weld runs, since the weld term is Rs 593 and the side-plate term is Rs 2667 - these dominate the total and are where a real saving exists. Report the saving against the foreman's sketch as the justification for the redesign.