Practice Problems: Module B, Linear Programming
Problem 1:
Chad’s Pottery Barn has enough clay to make 24 small vases or 6 large vases. He has only enough
of a special glazing compound to glaze 16 of the small vases or 8 of the large vases. Let X1 = the
number of small vases and X2 = the number of large vases.
The smaller vases sell for $3 each, and the larger vases would bring $9 each.
(a) Formulate the problem
(b) Solve the problem graphically
Problem 2:
A fabric firm has received an order for cloth specified to contain at least 45 pounds of cotton and
25 pounds of silk. The cloth can be woven out of any suitable mix of two yarns A and B. They
contain the proportions of cotton and silk (by weight) as shown in the following table:
Cotton
Silk
A
30%
50%
B
60%
10%
Material A costs $3 per pound, and B costs $2 per pound. What quantities (pounds) of A and B
yarns should be used to minimize the cost of this order?
ANSWERS
Problem 1:
(a) Formulation:
Objective function:
Subject to:
Maximize
3X1 + 9X2
Clay constraint:
1X1 + 4X2  24
Glaze constraint:
1X1 + 2X2  16
(b) Graphical Solution
X1 @ $3.00
X2 @ $9.00
Income
A
0
0
0
B
0
6
$54
C
8
4
$60*
D
16
0
$48
Evaluating all possible corner points that might be the optimal solution, the optimum income of $60
will occur by making and selling 8 small vases and 4 large vases. An iso-profit line on the graph
from (20,0) to (0,6.67) shows the point that returns value of $60.
Problem 2:
Formulation:
Objective function: min C =
Constraints:
3A +
2B
Cotton
.30A + .60B  45
Silk
.50A + .10B  25
We can learn the values of A and B at intersection of the Silk and Cotton constraints by
simultaneously solving the equations that determine the point. To solve for A we first multiply the
Silk equation by 6 then subtract the Cotton equation.
3.0 A  .60 B  150 (Silk constraint multiplied by 6)
.30 A  .60 B  45 (subtract Cotton equation)
2.70 A
 105
A
 38.8
Following the same basic procedure for the value of B, we multiply the Cotton equation by 3 and
the Silk equation by 5 and subtract the Silk equation.
1.50 A  3.0 B  225 (Cotton equation multiplied by 5)
1.50 A  .30 B  75 (Silk equation multiplied by 3 and subtracted)
2.70 B  150
B  55.6
Using the Objective Function, we can calculate the profit at each of the three corner points:
Axis intercept (0, 250) = (0 * $3) + (250 * $2) = $500
Axis intercept (150, 0) = (150 * $3) + 0 * $2) = $350
Intersection of the two constraints (38.8, 55.5) = (38.8 * $3) + (55.6 * $2) = $227.60
The minimum cost is found at the intersection of the two constraint equations.