1. ## Linear Programming Problem!!

Consider the following linear programming problem:

Maximise Z = 100x1 + 200x2

Subject to 3x1 + 2x2 ≤ 120
x1 + 4x2 ≤ 80
2x1 + x2 ≤ 100
x1, x2 ≥ 0

Solution is Z=5600
x1=32
x2=12
s3=24

If a new constraint 2x1+ 2x2 ≤ 100 was included in the original problem, how would this affect the solution?

My answer is that there is no change. But is this correct and is tehre a quicker way of getting the answer rather than the long drawn-out simplex method??

***EDIT I have another question where there are 3 variables X,Y, and Z. How do I plot that/answer the question then??? ***

2. Hello, DooBeeDoo!

Consider the following linear programming problem:

Maximise: . $Z \:= \:100x + 200y$

Subject to: . $\begin{Bmatrix}3x + 2y & \leq &120 & [1]\\ x + 4y & \leq & 80 & [2]\\ 2x + y & \leq & 100 & [3]\\ x, y & \geq & 0 & [4]\end{Bmatrix}$

Solution is: . $Z=5600,\;x =32,\;y=12,\;s3=24$

If a new constraint $2x+ 2y \leq 100$ was included in the original problem,
how would this affect the solution?

My answer is that there is no change. . . . . I agree

Is there a quicker way of getting the answer?
. . You weren't shown how to graph these problems?

Constraint [4] puts us in Quandrant 1.

For $[1]\;3x + 2y \:\leq \:120$, graph the line: . $3x + 2y \:=\:120$
It has intercepts: (40, 0), (0, 60)
Draw the line and shade the region below it.

For $[2]\;x + 4y \:\leq \:80$ graph the line: . $x + 4y \:=\:80$
It has intercepts: (80, 0), (0, 20)
Draw the line and shade the region below it.

The shaded region looks like this . . .
Code:
        |
60 *
|*
| *
|  *
|   *
|    *
20 o     *
|:::*  *
|:::::::o
|::::::::*  *
|:::::::::*     *
|::::::::::*        *
- - o - - - - - o - - - - - * - -
|          40          80

Constraint [3] is useless.
The line $2x + y \:=\:100$ has intercepts (50, 0), (0, 100)
. . and falls outside of the shaded region.

We are concerned with the vertices of the shaded region.
We can see three of them: .(0, 0), (40, 0), (0, 20)

We find the fourth by locating the intersection of the two slanted lines.

We solve the system: . $\begin{array}{ccc}3x + 2y &=&120 \\ x + 4y &=& 80\end{array}\;\text{ and get: }\32,\:12)" alt="\begin{array}{ccc}3x + 2y &=&120 \\ x + 4y &=& 80\end{array}\;\text{ and get: }\32,\:12)" />

We substitute these coordinates into the Z-function
. . to determine which one maximizes Z.

3. thanks a lot dude. very helpful.

4. I have another question where there are 3 variables X,Y, and Z. How do I plot that/answer the question then???

5. Hello, DooBeeDoo!

I have another question where there are 3 variables X,Y, and Z.
How do I plot that / answer the question then?

With three or more variables, there are mechanical procedures.
. . The most common is the Simplex Method.