SOLUTION: Consider the following final tableau corresponding to a linear programming problem. x y z u v w P Constants 0 0 −1/2 1 −1/4 −1/4

Algebra.Com
Question 1202966: Consider the following final tableau corresponding to a linear programming problem.
x y z u v w P Constants
0 0 −1/2 1 −1/4 −1/4 0 2
0 1 0 0 1/4 −1/4 0 3
1 0 3/2 0 0 1/2 0 15
0 0 1 0 3/2 1/2 1 78
Part 1 of 2
(a.) How many solutions does the linear programming problem have?
One Solution
Part 2 of 2
(b.) Since the linear programming problem has one solution, provide the solution as a point.
(x, y, z) = ( , , )

Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
How did you know there was only 1 solution if you didn't know how to do
the other part?  Did you just guess?



That is the matrix for this system of equations:



Solve the bottom equation for P:



We want P to be as large as possible, and we have three non-negative numbers
subtracted from the 78.  We can keep the whole 78 for P by choosing all three
variables z, v, and w = 0.  So we substitute those in the system and get:



So P has a maximum value of 78 when x = 15, y = 3, and z = 0.  The point is

(x,y,z) = (15,3,0)

Edwin

RELATED QUESTIONS

Consider the following final tableau corresponding to a linear programming problem. x (answered by Edwin McCravy)
Consider the following simplex tableau. x y z u v w P Constants 0 1/2 0 1 −1/2 (answered by Edwin McCravy)
Consider the following simplex tableau. x y u v P Constants 1 1 1 0 0 1 1 0 −1... (answered by Edwin McCravy)
Determine whether the given simplex tableau is in final form. If so, find the solution to (answered by stanbon)
The initil tableau of a linear programming problem is given below. Use the simplex... (answered by Edwin McCravy)
In solving a linear programming problem, you are given the following initial tableau. (answered by Edwin McCravy)
1:use the simple method to solve the following linear programming (L.P.P) Maximize... (answered by Theo)
Solve the linear programming problem. Minimize z = 11x + 6y + 7 subject to: x ≥ 0,... (answered by ikleyn)
x1 x2 x3 s1 s2 z 3 3 2 1 0 0 12 2 1 2 0 1 0 14 -1 -1 -3 (answered by richwmiller)