A numerical method which is more efficient than repeated subtabulation for obtaining the optimal solution is the following bracketing method. The initial tabulation locates an interval in which the solution occurs. The optimal solution is then estimated by optimizing a suitable quadratic approximation. Consider again the milk carton problem, Example 8.35. Calculate A(70), A(80) and A(90) and deduce that a minimum occurs in [70, 90]. Next find numbers p, q and r such that

satisfies C(70) = A(70), C(80) = A(80) and C(90) = A(90). The minimum of C occurs at 80 – q/(2p). Show that this yields the estimate b = 82.2. Evaluate A(82.2) and deduce that the solution lies in the interval [80, 90]. Next repeat the process using the values A(80), A(82.2) and A(90) and show that the solution lies in the interval [80, 83.1]. Apply the method once more to obtain an improved estimate of the solution.

Data from Example 8.35

A milk retailer wishes to design a milk carton that has a square cross-section, as illustrated in Figure 8.39(a), and is to contain two pints of milk (2 pints ≡ 1.136 litres). The carton is to be made from a rectangular sheet of waxed cardboard, by folding into a square tube and sealing down the edge, and then folding and sealing the top and bottom. To make the resulting carton airtight and robust for handling, an overlap of at least 5 mm is needed. The procedure is illustrated in Figure 8.39(b). As the milk retailer will be using a large number of such cartons, there is a requirement to use the design that is least expensive to produce. In particular the retailer desires the design that minimizes the amount of waxed cardboard used.

Transcribed Image Text:

C(b) p(b80) + q(b 80) + r =