Chapter 6.6, Problem 1P

Explanation of Solution

Duals of LP:

• Consider the linear programming problem given below:
• max z= $3x_1+2x_2$
• Subject to constraints:
  • $\frac{1}{40}{x}_1+\frac{1}{60}{x}_2\le 1$
  • $\frac{1}{50}{x}_1+\frac{1}{50}{x}_2\le 1$
  • $x_1,x_2\ge 0$
• To find the dual of the above LP proceed as follows:

Max z
min w	($x_1\ge 0$)	($x_2\ge 0$)
	$x_1$	$x_2$
$\left(y_1\ge 0\right)$	$y_1$	$\frac{1}{40}$	$\frac{1}{60}$	$\le 1$
$\left(y_2\ge 0\right)$	$y_2$	$\frac{1}{50}$	$\frac{1}{50}$	$\le 1$
Profit (in 100s of dollars)	$\ge 3$	$\ge 2$

• The dual is:
• min w=$y_{}$