please help cant solve linear programming problem

jorge

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you have been hired to manage a tire production plant you will oversee two production processes each prcess involves the use of electricians operators and machines under the currenct contract the electricians can work a total of 8400 hours the operators that run the machines can work a total of 4800 hours therefor limiting the machine time to 4800 hours use the chart below to help you figure out how many tires from each process you should make to maxamize your profit

hours to make one tire process A process B
elctricians 6 6
operators 2 4
machine time 4 2
profit per tire 40$ 50$


the paper im working off of says the answer is 400 tires from process A and 1000 from process B but i cant figure out how they got those numbers =( please help
 
you have been hired to manage a tire production plant you will oversee two production processes each prcess involves the use of electricians operators and machines under the currenct contract the electricians can work a total of 8400 hours the operators that run the machines can work a total of 4800 hours therefor limiting the machine time to 4800 hours use the chart below to help you figure out how many tires from each process you should make to maxamize your profit

hours to make one tire process A process B
elctricians 6 6
operators 2 4
machine time 4 2
profit per tire 40$ 50$


the paper im working off of says the answer is 400 tires from process A and 1000 from process B but i cant figure out how they got those numbers =( please help
What you tried so far? What are your variables? What is your objective function? What are your constraints, including the non-negativity constraints?
 
my constraints

What you tried so far? What are your variables? What is your objective function? What are your constraints, including the non-negativity constraints?
my objective function is

p=40x+50y

constraints are

6x+6y less than or equal to 8400

2x+4yless than or equal to 4800 for the operators

4x+2y less than or equal to 4800 for the machine time

x greater than or equal to 0

y greater than equal to zero

are my constraints thus far =(
 
my objective function is

p=40x+50y

constraints are

6x+6y less than or equal to 8400

2x+4yless than or equal to 4800 for the operators

4x+2y less than or equal to 4800 for the machine time

x greater than or equal to 0

y greater than equal to zero

are my constraints thus far =(
Perfect.

Now here is the central idea. Consider any (a, b) such that x >= 0 and y >= 0
but 6x + 6y < 8400, 2x + 4y < 4800, and 4x + 2y < 4800. We can increase the objective function by increasing either x or y until one of the less-than constraints is reached. We then can improve the objective function more by moving along that constraint until we meet another constraint and so on. We check every intersection of constraints for feasibility and optimality.

Let's give it a go.

Choose x = 500 = y. The objective function has a value of
40 * 500 + 50 * 500 = 20000 + 25000 = 45000.

By unit, B is more profitable so let's reduce x to zero and increase y as much as possible. If x is zero, then the constraints individually say that y could be as high as 2400, 1400, or 1200. But we must satisfy all three constraints so 1200 for y is the best we can do if we reduce x to 0. That increases the objective function to 1200 * 50 = 60000. The point (0, 1200) is on TWO constraint lines,
x = 0 and 2x + 4y = 4800.

But we can do better. By giving up 1 y and adding 2 x, we stay on the constraint line but increase the objective function by 2 * 40 - 50 = 30.

Eventually we meet the 6x + 6y = 8400 constraint when x = 400 and y = 1000.

Notice that 2 * 400 + 4 * 1000 = 4800. So again we are on TWO constraint lines.

The objective function is 40 * 400 + 50 * 1000 = 16000 + 50000 = 66000,
which is better than 60000.

How about the other constraints. 400 > 0. 1000 > 0.

And 4 * 400 + 2 * 1000 = 3600 < 4800. All feasible

So the technique is to look at ALL intersections of constraint lines. Test each intersection for feasibility and optimality. Finally choose the optimal feasible intersection.
 
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