Question

Find the maximum possible value of f = 80x + 160y and the values of x and y that give that value, subject to the constraints. 3x + 4y = 28 x + 2y = 13 2x+ys17 x20, y20 Select the correct choice below and fill in any answer boxes within your choice. O A. The maximum value is f= at the point . (Type integers or decimals. Type an ordered pair.) OB. The maximum value is f= at all points on the line segment from the point (Type integers or decimals. Type ordered pairs.) on the left to the point on the right.

Answer 1

we need to maximize the function , $f=80x+60y$

subject to ,

3x + 4y < 28

x +2y < 13

$2x+y \leq 17$

by adding slack variables to inequality we get the initial system as

$3x+4y+s_1=28$

$x+2y+s_2=13$

$2x+y+s_3=17$

$f-80x-60y=0$

we put this in a table to obtain the initial simplex tableau ,

$\boldsymbol{x}$	$\boldsymbol{y}$	$\boldsymbol{s_1}$	$\boldsymbol{s_2}$	$\boldsymbol{s_3}$	$\boldsymbol{f}$	$\boldsymbol{RHS}$
3	4	1	0	0	0	28
1	2	0	1	0	0	13
2	1	0	0	1	0	17
-80	-160	0	0	0	1	0

Selecting a pivot column which will be the column containing the largest negative coefficient in objective function [last row ] it is -160 which is column y .

Now we need to select pivot row by selecting the row with smallest test ratio . [=RHS / coeff. of y]

for row 1 we get , test ratio = 28/4 = 7

for row 2 we get test ratio = 13/2 = 6.5

for row 3 we get test ratio = 17/1 = 17

so we get row 2 as pivot row ,

consider our matrix form as ,

$\begin{bmatrix} 3 & 4 & 1 & 0 & 0 & 0 &|&28 \\ 1 &2 &0 &1 & 0 & 0 & |&13\\ 2& 1 &0 &0 &1 &0 &|&17 \\ -80 &-160 & 0 &0 &0 &1&| &0 \end{bmatrix}$

to eliminate row 1 , 3 , 4 we do row transformation as

$R_2 \to R_2/2$

$\begin{bmatrix} 3 & 4 & 1 & 0 & 0 & 0 &|&28 \\ 1/2 &1 &0 &1/2 & 0 & 0 & |&13/2\\ 2& 1 &0 &0 &1 &0 &|&17 \\ -80 &-160 & 0 &0 &0 &1&| &0 \end{bmatrix}$

$R_1 \to R_1-4R_2$

$R_3 \to R_3-R_2$

$R_4 \to R_4+160R_2$

$\begin{bmatrix} 1 & 0 & 1 & -2 & 0 & 0 &|&2 \\ 1/2 &1 &0 &1/2 & 0 & 0 & |&13/2\\ 3/2&0 &0 &-1/2 &1 &0 &|&21/2 \\ 0 &0 & 0 &80 &0 &1&| &1040 \end{bmatrix}$

so we get optimal solution at x = 0 , y = 13/2 [second row ]

and maximum value is 1040

the maximum value of f is 1040 at the point (0,6.5)