Linear Programming - Elimination Method help

Hi: This may be slightly more advanced than basic algebra but I think it fits.

I have the following equation:

**Maximize Z = 60x + 90y**

Subject to:

7x+7y < or = 350

160x+80y > or = 4000

y < or = 25

x > or = 20

This is a maximization / linear programming example where I'd find the feasable region and then determine the optimum solution.

My first step was to sub 0 in for x and y in both equations, getting 25,0 and 50,0 and 0,50 for my coordinates. The constraints of y<or =20 and x > or =0 give me the feasable region but I am having trouble using the substution method to solve for the optimum solution.

Every example I have seen and looked up have been simple numbers where you can either add down, subtract, or multiple by a small amount to get a variable to cancel out. In this situation, 7 does not go into 80 or 160 evenly.

I tried using the LCM and came up with 80, but when I run that number through I end up with 560x = 0, so I know I must be doing something wrong.

How do you handle numbers that do not cancel out evenly?

Thanks!!

Re: Linear Programming - Elimination Method help

Quote:

Originally Posted by

**ljj** Hi: This may be slightly more advanced than basic algebra but I think it fits.

I have the following equation:

**Maximize Z = 60x + 90y**

Subject to:

7x+7y < or = 350

160x+80y > or = 4000

y < or = 25

x > or = 20

This is a maximization / linear programming example where I'd find the feasable region and then determine the optimum solution.

My first step was to sub 0 in for x and y in both equations, getting 25,0 and 50,0 and 0,50 for my coordinates. The constraints of y<or =20 and x > or =0 give me the feasable region but I am having trouble using the substution method to solve for the optimum solution.

Every example I have seen and looked up have been simple numbers where you can either add down, subtract, or multiple by a small amount to get a variable to cancel out. In this situation, 7 does not go into 80 or 160 evenly.

I tried using the LCM and came up with 80, but when I run that number through I end up with 560x = 0, so I know I must be doing something wrong.

How do you handle numbers that do not cancel out evenly?

Thanks!!

Hint: 7 divides 350! The first equation reduces to

$\displaystyle x+y \le 50$

Now can you use it for elimination?

Re: Linear Programming - Elimination Method help

Hello, ljj!

Quote:

$\displaystyle \text{Maximize: }z \:=\: 60x + 90y$

$\displaystyle \text{Subject to: }\:\begin{Bmatrix}x+y \:\le \:50 & [1] \\ 2x+y \:\ge\:50 & [2] \\ x \ge 20 & [3] \\ y \le 25 & [4]\end{Bmatrix}$

The line of [1] has intercepts (50, 0) and (0, 50).

Draw the line and shade the region below the line.

The line of [2] has intercepts (25,0) and (0,50).

Draw the line and shade the region above the line.

The line of [3] is the vertical line: $\displaystyle x \,=\,20.$

Draw the line and shade the region to the right of the line.

The line of [4] is the horizontal line: $\displaystyle y \,=\,25.$

Draw the line and shade the region below the line.

The critical region looks like this:

Code:

` |`

*

|**

| * * |

| * * |

| * *

| * D| * C

- + - - * o---o

| *|:::::*

| Eo:::::::*

| |*::::::::*

- - + - - - + o---------o - -

| A B

The vertices are: .$\displaystyle \begin{Bmatrix}A:& (25,0) \\ B: & (50,0) \\ C: & (25,25) \\ D: ^& (20,25) \\ E: & (20,10) \end{Bmatrix}$

Test the vertices in the *z*-function and determine the maximum *z.*