4. Find if possible, the minimum value of the objective function
3x - 4y subject to the constraints
- 2x + y $12, x - y s2, x 20 and y 20.
(a) 0
(b) - 36
(c) 8
(d) No solution
Answers
Answer:
SOLUTION OF LINEAR PROGRAMMING PROBLEMS
THEOREM 1 If a linear programming problem has a solution, then it must occur at a vertex, or corner
point, of the feasible set, S, associated with the problem. Furthermore, if the objective function P is
optimized at two adjacent vertices of S, then it is optimized at every point on the line segment joining
these two vertices, in which case there are infinitely many solutions to the problem.
THEOREM 2 Suppose we are given a linear programming problem with a feasible set S and an objective
function P = ax+by. Then,
If S is bounded then P has both a maximum and minimum value on S
If S is unbounded and both a and b are nonnegative, then P has a minimum value on S provided
that the constraints defining S include the inequalities x≥ 0 and y≥ 0.
If S is the empty set, then the linear programming problem has no solution; that is, P has neither
a maximum nor a minimum value.
THE METHOD OF CORNERS
Graph the feasible set (region), S.
Find the EXACT coordinates of all vertices (corner points) of S.
Evaluate the objective function, P, at each vertex
The maximum (if it exists) is the largest value of P at a vertex. The minimum is the smallest value
of P at a vertex. If the objective function is maximized (or minimized) at two vertices, it is
minimized (or maximized) at every point connecting the two vertices.
Example: Find the maximum and minimum values of P=3x+2y subject to
x + 4y ≤ 20
2x + 3y ≤ 30
x≥0 y≥ 0
1. Graph the feasible region.
Start with the line x+4y=20. Find the intercepts by letting x=0 and y=0 : (0,5) and (20,0).
Test the origin: (0) + 4(0) ≤ 20 . This is true, so we keep the half-plane containing the origin.
Graph the line 2x+3y=30. Find the intercepts (0,10) and (15,0).
Test the origin and find it true. Shade the upper half plane.
Non-negativity keeps us in the first quadrant.
2. Find the corner points. We find (0,0) is one corner, (0,5) is the corner from the y-intercept of the first
equation, The corner (15,0) is from the second equation. We can find the intersection of the two lines
using intersect on the calculator or using rref: Value is (12, 2)
3. Evaluate the objective function at each vertex. Put the vertices into a table:
Vertex P=3x+2y
(0, 0) 0 min
(0, 5) 10
(15, 0) 45
(12, 2) 40 Max
4. The region is bounded, therefore a max and a min exist on S. The minimum is at the point (0,0) with a
value of P=0 . The maximum is at the point (15,0) and the value is P=45 .
Step-by-step explanation:
Example: A farmer wants to customize his fertilizer for his current crop. He can buy plant food mix A
and plant food mix B. Each cubic yard of food A contains 20 pounds of phosphoric acid, 30 pounds of
nitrogen and 5 pounds of potash. Each cubic yard of food B contains 10 pounds of phosphoric acid, 30
pounds of nitrogen and 10 pounds of potash. He requires a minimum of 460 pounds of phosphoric acid,
960 pounds of nitrogen and 220 pounds of potash. If food A costs $30 per cubic yard and food B costs
$35 per cubic yard, how many cubic yards of each food should the farmer blend to meet the minimum
chemical requirements at a minimal cost? What is this cost?
Before we can use the method of corners we must set up our system. Start by defining your variables
properly and completely:
x = no. of cubic yards of plant food mix A
y = no. of cubic yards of plant food mix B
Next find the objective function. This is the quantity to be minimized or maximized:
Minimize C=30x+35y
Finally, set up the constraints on the problem:
Subject to
20x + 10y ≥ 460 phosphoric acid requirement
30x + 30y ≥ 960 nitrogen requirement
5x + 10y ≥ 220 potash requirement
x≥0 y≥ 0
Now use the method of corners:
1. Graph the feasible region. Start by putting in the three lines and finding their intercepts. Test the origin
in each inequality and find that the origin is false, so we shade the lower half-plane of each. This feasible
region is unbounded. That means that there is a minimum, but no maximum.
2. Find the corner points. We have 4 corners of the feasible region.
3. Evaluate the objective function at each vertex. Put the vertices into a table:
Vertex C=30x+35y
(0, 46) 1610
(44, 0) 1320
(20, 12) 1020 minimum
(14, 18) 1050
4. The minimum occurs at (20,12). So the farmer should buy 20 cubic yards of plant food mix A and 12
cubic yards of plant food mix B at a cost of $1020.
Answer:
The minimum value of 3x - 4y is -48.
Step-by-step explanation:
Given 3x - 4y subject to the constraints -2x + y 12, x - y
2, x
0 and y
0
Let Z = 3x - 4y
At O(0,0) , Z = 3(0)-4(0) = 0
A(0,12) , Z = 3(0)-4(12) = -48
B(2,0) , Z = 3(2)-4(0) = 6
Hence minimum value of Z= -48