Question

(a) Newbury Fastenings produce two kinds of specialized machine components, S10X and S10Y for which demand exceeds capacity. The production costs for the two products are given in the table below:

S10X S10Y
per component per component
Materials in Kg
(at £8 per Kg) 3 2

Labour in hours
(at £7 per hour) 2 3

Other Variable Costs (£) 6 5


The selling prices of the two products are £79 per unit and £67 per unit respectively. The production manager has stated at the monthly review meeting that there will only be 4,500 kg of material and 4,000 labour hours available next month. The company works on Just-in-Time manufacturing so holds no stocks and it wishes to maximize its profit.
How should the manufacturer arrange the production for next month in order to maximize contribution to profit? Determine the optimum level of contribution to profit.

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c} & S10X & S10Y \\ \hline \begin{array}{cc} Materials\ in\ kg \\ (at\ £8\ per\ Kg) \end{array} & 3 & 2 \\ \hline \begin{array}{cc} Labour\ in\ hours \\ (at\ £7\ per\ hour) \end{array} & 2 & 3 \\ \hline \begin{array}{cc} Other\ Variable\ Costs \\ ( £ ) \end{array} & 6 & 5 \end{array}$$

Let $X=$ the number of S10X products, $Y=$ the number of S10Y products.

$$3X+2Y\leq4500$$ $$2X+3Y\leq4000$$

$$P=(79-3\cdot8-2\cdot 7-6)X+(67-2\cdot8-2\cdot 7-5)Y=$$

$$=35X+32Y$$

$$X=0: 0\leq Y \leq{4000\over 3}, P_{max}=42666.67$$

$$Y=0: 0\leq X \leq1500, P_{max}=52500$$

$$Y_1=-{3\over 2}X+2250$$

$$Y_2=-{2\over 3}X+{4000\over 3}$$

$$Y_1=Y_2:-{3\over 2}X+2250=-{2\over 3}X+{4000\over 3}$$

$${5\over 6}X={2750\over 3}$$

$$X=1100$$

$$0\leq X\le 1100: Y=-{2\over 3}X+{4000\over 3}$$

$$P=35X-{64\over 3}X+{128000\over 3}={41\over 3}X+{128000\over 3}$$

$$P_{max}={41\over 3}(1100)+{128000\over 3}=57700$$

$$1100\leq X\le 1500: Y=-{3\over 2}X+2250$$

$$P=35X-48X+72000=-13X+72000$$

$$P_{max}=-13(1100)+72000=57700$$

If $X=1100,Y=600,$ we have the maximum profit

$$P_{max}=35(1100)+32(600)=£\ 57700$$