Linear Programming using Graphing Calculator

In summary: So the problem is that the $X$ variable is not in the correct position.I suspect your daughter is calculating something different from what was intended.
  • #1
safaulk
1
0
My daughter needs serious help with Linear Programming using Graphing Calculator. She has a problem she has been trying to solve for over a month, I have tried to help her and searched endlessly for help. Hopefully, the link below works. Everything comes up fine except for the very last part. We keep getting either an Error message or 137 instead of 78, as the problem shows. Thanks in advance for any help you can provide.

https://www.dropbox.com/s/xjq0i38dz8j6jeh/Photo Nov 16, 1 05 11 PM.jpg
 
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  • #2
Welcome to the forum.

I can't help you with using the calculator (perhaps somebody else on this forum can), but I can explain why the answer is 136, as the linked picture says. If you are getting 137, that's pretty close to the correct result. Why are you saying that you are expected to get 78?

For starters, here are the relevant graphs (the picture is clickable).

[GRAPH]00wufkkqgc[/GRAPH]

Please let me know if you need help of this sort.
 
  • #3
safaulk said:
My daughter needs serious help with Linear Programming using Graphing Calculator. She has a problem she has been trying to solve for over a month, I have tried to help her and searched endlessly for help. Hopefully, the link below works. Everything comes up fine except for the very last part. We keep getting either an Error message or 137 instead of 78, as the problem shows. Thanks in advance for any help you can provide.

https://www.dropbox.com/s/xjq0i38dz8j6jeh/Photo Nov 16, 1 05 11 PM.jpg

I suspect your daughter is calculating something different from what was intended.
Let me try and explain the reason for the answer 78 and perhaps it will help you two to figure out what whent wrong.

The answer 78 should come out for the bottom right corner of the double shaded region.
This is where the line $Y=8X-48$ has its ZERO ($Y = 0$).
In other words, where $8X-48 = 0$.
The corresponding $X = 6$.
So the bottom right corner is at $X = 6$ and $Y = 0$.

The objective function in the bottom right corner is
$$13X + 2Y = 13 \times 6 + 2 \times 0 = 78$$
 

Related to Linear Programming using Graphing Calculator

1. What is linear programming using a graphing calculator?

Linear programming is a mathematical method used to optimize a linear objective function, subject to linear equality and inequality constraints. A graphing calculator can be used to graph the constraints and visually determine the optimal solution.

2. How do I enter constraints into a graphing calculator for linear programming?

To enter constraints into a graphing calculator, first graph each constraint as an equation. Then, use the calculator's "Inequalz" feature to shade the feasible region. Finally, use the "Max" or "Min" function to find the optimal solution.

3. Can a graphing calculator handle multiple constraints for linear programming?

Yes, a graphing calculator can handle multiple constraints for linear programming. It can graph and shade the feasible region for multiple constraints, and use the "Max" or "Min" function to find the optimal solution.

4. Are there any limitations to using a graphing calculator for linear programming?

While a graphing calculator can handle most linear programming problems, there are some limitations. It may not be able to handle a large number of constraints or variables, and it cannot handle nonlinear functions or constraints.

5. How do I interpret the results from a graphing calculator for linear programming?

The results from a graphing calculator for linear programming will include the optimal solution, which is the point where the objective function is either maximized or minimized. It will also show the coordinates of this point, as well as the values of the decision variables that make up this point.

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