Question

A small business enterprise makes dresses and trousers. To make a dress

requires 1

2

hour of cutting and 20 minutes of stitching. To make a trouser

requires 15 minutes of cutting and 1

2

hours of stitching. The profit on a dress is

P40 and on a pair of trousers is P50. The business operates for a maximum of

8 hours per day. Determine how many dresses and trousers should be made to

maximize profit and what the maximum profit is.

Answer 1

Let x,y- number of dresses and trousers to done.

Mathematical model is

z=40$\cdot$ x+25$\cdot$ y$\rarr$ max;

50$\cdot$ x+45$\cdot$ y$\le$ 480

(or 10$\cdot$ x+9$\cdot$ y$\le$ 96)

x,y$\ge$ 0

x,y $\in$ Z

Because x is integer $x\in \{0,1,2,3,4,5,6,7,8,9\}$

1) x=9,$y=[\frac{96-9\cdot 10}{9}]=[0.6]=0;z=40\cdot 9+25\cdot 0=360$

2) x=8,y=$\frac{96-8\cdot 10}{9}]=[1 \frac{7}{9}]=1;z=40\cdot 8+25\cdot 1=345$

3) x=7,y=$\frac{96-7\cdot 10}{9}]=[2 \frac{8}{9}]=2;z=40\cdot 7+25\cdot 2=330$

4) x=6,y=$\frac{96-6\cdot 10}{9}]=[4]=4;z=40\cdot 6+25\cdot 4=340$

5) x=5,y=$\frac{96-5\cdot 10}{9}]=[5 \frac{1}{9}]=5;z=40\cdot 5+25\cdot 5=325$

6) x=4,y=$\frac{96-4\cdot 10}{9}]=[6 \frac{2}{9}]=6;z=40\cdot 4+25\cdot 6=310$

7) x=3,y=$\frac{96-3\cdot 10}{9}]=[7 \frac{2}{9}]=7;z=40\cdot 3+25\cdot 7=295$

cases x=0,1,2 worse than dest case (x=10) because 2$\cdot 40+10\cdot 25=330<360$

Now all possibilities are considered and optimal solution (x_opt,y_opt)=(10,0) with z_opt=360.

So x_opt=10,y_opt=0,z_opt=360.