14) Find the value of x and y where P = 12x + 14y is maximum.
3x + 3y = 8
4y + 2x 58
Answers
Step-by-step explanation:
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 avalue of P=0 . The maximum is at the point (15,0) and the value is P=45 .this in slope-intercept form, y = -1.25x + 50 and 120/7 ≤ x ≤ 40 .Example: Find the maximum value of f=10x+8y subject to
2x + 3y ≤ 120
5x + 4y ≤ 200
x≥0 y≥ 0
1. Graph the feasible region.
2x+3y=120 Intercepts are (0,40) and (60,0). Origin is true so shade the upper half plane.
5x+4y=200 Intercepts are (0,50) and (40,0). Origin is true so shade the upper half plane.
2. Find the corner points. We find (0,0) is one corner, (0,40) is the corner from the y-intercept of the first
equation. The corner (40,0) is from the second equation. We can find the intersection of the two lines
using the intersection or rref: (120/7, 200/7).
3. Evaluate the objective function at each vertex. Put the vertices into a table:
Vertex F=10x+8y
(0, 0) 0
(0, 40) 320
(40, 0) 400 Max
(120/7, 200/7) 400 Max
4. We find that the objective function is maximized at two points, therefore it is maximized along the line
segment connecting these two points. The value of the objective function is 400 = f = 10x + 8y. Writecubic yards of plant food mix B at a cost of $1020.
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