Chapter 3: Section 3-3
Solutions of Linear Programming Problems
D. S. Malik
Creighton University, Omaha, NE
Geometric Approach to Solve Linear Programming (LP) Problems
The objective of a linear programming problem is to maximize or minimize an objective function, subject to certain constraints.
We have considered several examples illustrating how to formulate the constraints and the objective function.
In the constraints, the inequalities use either the symbol or the symbol , i.e., none of the inequalities are strict.
This is important because the …rst step in solving a linear
programming problem using the geometric approach is to draw the graphs of the inequalities and determine the region (solution set) that satis…es the linear inequalities.
Next determine the corner points of the solution set.
If an inequality is strict, then a corner point given by that inequality may not be in the solution set.
Thus, in all of the linear programming problems that we consider, the
linear inequalities will not be strict.
Feasible Region
Feasible Region (Solution Set)
BoundedUnbounded
Bounded Feasible Region
Theorem
( Bounded Region) Let F be the feasible region of a linear programming problem and F 6= ∅. Let f be the objective function of the linear
programming problem. Suppose that F is bounded. Then
(i) f attains its maximum value at a corner point of F ,
(ii) f attains its minimum value at a corner point of F .
Procedure to …nd a solution of a linear programming problem when the feasible region is bounded.
1
Graph the linear inequalities and determine the feasible region (solution set).
2
Find the corner points of the feasible region.
3
Evaluate the objective function at each of the corner points.
4
If the problem is to maximize the objective function, then choose the largest value of the objective function. The coordinates of the corner point that gives the largest value is a solution of the linear
programming problem. (If there is more than one such corner point, then choose one of those points.)
5
If the problem is to minimize the objective function, then choose the
smallest value of the objective function. The coordinates of the
corner point that gives the smallest value is a solution of the linear
programming problem. (If there is more than one such corner point,
then choose one of those points.)
Example
Consider the following constraints of a linear programming problem:
x 3, x + y 0, x y 0, x + 2y 5.
The region in the x-y plane that satis…es these inequalities is:
x y
x + y = 0 x - y = 0
A(-(5/3),(5/3))
B(3, 4) C(3,3)
-x + 2y = 5
-5
x = 3
O(0,0) 1 2 3 4
-1 -2
0 1 2 3
-1 -2
-3 4 5
-4
Corner points of the feasible region are: O ( 0, 0 ) , A
53,
53, B ( 3, 4 ) ,
and C ( 3, 3 ) .
Example
The corner points of the feasible region are: O ( 0, 0 ) , A
53,
53, B ( 3, 4 ) , and C ( 3, 3 ) .
Let us minimize 6x 3y subject to the given constraints.
Because the feasible region is bounded, the minimum occurs at a corner point. Let us evaluate 6x 3y at the corner points. Now
Corner Point Value of the Objective function: 6x 3y O ( 0, 0 ) 6 0 3 0 = 0
A
53,
356 (
53) 3
53= 10 5 = 15 B ( 3, 4 ) 6 3 3 4 = 18 12 = 6
C ( 3, 3 ) 6 3 3 3 = 18 9 = 9
The minimum value of 6x 3y is 15 and it occurs at A
53,
53.
Example
The corner points of the feasible region are: O ( 0, 0 ) , A
53,
53, B ( 3, 4 ) , and C ( 3, 3 ) .
Let us maximize 5x + 2y subject to the given constraints.
Because the feasible region is bounded, the maximum occurs at a corner point. Let us evaluate 5x + 2y at the corner points. Now
Corner Point Value of the Objective function: 5x + 2y O ( 0, 0 ) 5 0 + 2 0 = 0
A
53,
535 (
53) + 2
53=
253+
103=
153B ( 3, 4 ) 5 3 + 2 4 = 15 + 8 = 23
C ( 3, 3 ) 5 3 + 2 3 = 15 + 6 = 21
The maximum value of 5x + 2y is 23 and it occurs at B ( 3, 4 ) .
Exercise: Consider the following exercise described earlier.
A pet store specializes in cats and bunnies. Each cat costs $9 and each bunny costs $6. The pro…t on each cat is $12 and on each bunny is $9.
The store cannot house more than 30 animals and cannot spend more than $216 to buy the pets. Under a special agreement the pet store must house at least 2 cats. How many pets of each type should be housed to maximize the pro…t?
Let x be the number of cats and y be the numbers of bunnies.
The linear programming problem describing this problem is: The linear programming problem is:
maximize: 12x + 9y subject to: x 2, y 0
x + y 30
3x + 2y 72
The graph of the inequalities x 2, y 0, x + y 30, 3x + 2y 72 is:
x y
-10 0 10 20 30
3x + 2y = 72 x + y = 30
40
10 20 30 40
C(12, 18) B(2,28)
A(2,0) x = 2
D(24, 0)
The corner points are A ( 2, 0 ) , B ( 2, 28 ) , C ( 12, 18 ) , D ( 24, 0 ) .
The corner points are A ( 2, 0 ) , B ( 2, 28 ) , C ( 12, 18 ) , D ( 24, 0 ) . Example
Let us evaluate 12x + 9y at the corner points. Now
Corner Point Value of the Objective function: 12x + 9y A ( 2, 0 ) 12 2 + 9 0 = 24
B ( 2, 28 ) 12 2 + 9 28 = 276 C ( 12, 18 ) 12 12 + 9 18 = 306 D ( 24, 0 ) 12 24 + 9 0 = 288
The maximum value of 12x + 9y is 306 and it occurs at C ( 12, 18 ) .
To maximize the pro…t, the pet store should house 12 cats and 18
bunnies.
Unbounded Feasible Set
Theorem
( Unbounded Feasible Region) Let F be the feasible region of a linear programming problem and F 6= ∅. Let f be the objective function of the linear programming problem. Suppose that F is unbounded.
(i) If f attains its maximum value, then the maximum value occurs at a corner point of F ,
(ii) If f attains its minimum value, then the minimum value occurs at a
corner point of F ..
Unbounded Feasible Set
Remark
If the feasible region is bounded, then the preceding theorem does not guarantee that an objective function will have a maximum or minimum. It only guarantees that, if the maximum or minimum occurs, then it would occur at a corner point. So how do we know that a maximum or minimum will occur? To give a general answer to this problem, we will state a theorem at the end of this section. However, there are some special cases.
Let us consider some special cases that you will encounter frequently when
solving a real-world problem.
Suppose that the feasible region is restricted to the …rst quadrant and it extends inde…nitely in the …rst quadrant. Consider the object function 2x + 3y . Because the objective function extends inde…nitely in the …rst quadrant, the value of 2x + 3y increases without bounds when the values of x and y increase in the feasible region. That is, 2x + 3y has no …xed largest value. So 2x + 3y has no maximum in the feasible region. In fact, if the objective function is ax + by , where a 0, b 0, and both a and b are not zero, then the objective function ax + by has no maximum.
Suppose that the feasible region is restricted to the …rst quadrant and
it extends inde…nitely in the …rst quadrant. However, the objective
function, say 3x + 5y , is to be minimized. In this case, the minimum
will occur at a corner point, because the value of 3x + 5y cannot be
made arbitrarily small. Thus, in general, in this case, if the objective
function is ax + by , where a 0, b 0, and both a and b are not
zero, then the objective function ax + by has the minimum, and the
minimum will occur at a corner point.
Suppose that the feasible region is restricted to the third quadrant
and it extends inde…nitely in the third quadrant. Suppose that the
objective function is ax + by , where a 0, b 0, and both a and b
are not zero. Then ax + by has the maximum, and the maximum will
occur at a corner point. Furthermore, ax + by has no minimum.
Example
A hiker wants to take a snack mix of peanuts and raisins.
The hiker wants 1000 calories and 120 grams of fat from the mix.
Each gram of peanuts contains 8 calories and 0.4 grams of fat, and costs 7 cents.
Each gram of raisins contains 2 calories and 0.6 grams of fat, and costs 4 cents.
How many grams of each food should the hiker take so that the cost of the snack is the minimum?
Suppose that the hiker takes x grams of peanuts and y grams of raisins.
Then
Variable Calories Fat Cost
Each gram of peanuts x 8 0.4 7
Each gram of raisins y 2 0.6 4
Total (required) 1000 120
The linear programming problem is:
minimize: 7x + 4y subject to
x 0
y 0
8x + 2y 1000
0.4x + 0.6y 120.
x 0, y 0, 8x + 2y 1000, 0.4x + 0.6y 120.
We graph these inequalities and obtain the following graph:
100
0 100 200 300 400 500
200
4x + y = 500 y
400
300 500
600 700 x A(0, 500)
C(300, 0) B(90, 140)
600
2x + 3y = 600
