The following is the mathematical model of a linear programming problem for profit:
Maximize subject to
Z = 2X1 + 3X2
4X1+9X2 ≤ 72 10X1 + 11X2 ≤ 110 17X1 + 9X2 ≤ 153
X1 , X2 ≥ 0
The constraint lines have been graphed below along with one
example profit line (dashed). The decision variable X1 is used as
the X axis of the graph. Use this information for questions
19 through 23.
A). Which of the following gives the constraint line that cuts the X2 axis at X2 = 17?
A. B. C. D. E.
Z = 2X1 + 3X2 4X1+9X2≤ 72
10X1 + 11X2 ≤ 110 17X1 + 9X2 ≤ 153
X1 , X2 ≥ 0
B). The example profit line cuts the X1 axis at X1 = 9. What is the example profit amount?
3
6
9
18
None of the above
Y coordinate plane has four quadrants to its graph. Here is shown only the first
C). The X,
quadrant (positive X, positive Y). Which of the following
determines this?
A. B. C. D. E.
A. B. C. D. E.
Z=2X1+3X2 4X1+9X2≤ 72 10X1 + 11X2 ≤ 110 17X1 + 9X2 ≤ 153
X1 , X2 ≥ 0
D). The optimal solution is X1 = 4.3 and X2 = 6.1. What is the
maximum total profit?
8.6
10.4
18.3
26.9
None of the above
E). At the
constraint 17X1 + 9X2 ≤ 153 is
optimal solution given in question 22 above, the value of the slack variable for the
-25
0
25
14.4
None of the above
I guess the sequence of questions and their options are not right.
I have solved on eexample in r
> library(lpSolve)
> f.obj <- c(2, 3)
> f.con <- matrix(c(4, 9,
+ 10, 11,
+ 17, 9), nrow = 3, byrow = TRUE)
> f.dir <- c("<=",
+ "<=",
+ "<=")
> f.rhs <- c(72,
+ 110,
+ 153)
> lp("max", f.obj, f.con, f.dir, f.rhs)
Success: the objective function is 26.86957
> lp("max", f.obj, f.con, f.dir, f.rhs)$solution
[1] 4.304348 6.086957
Hence the answer of question D is 26.9
E )
optimal solution given in question 22 above, the value of the slack variable for the -25
lp("max", c(1,1),f.con, c("<=", "<=", "="), c(72,110, -25))$solution is 0
Get Answers For Free
Most questions answered within 1 hours.