I am doing some linear programing and was woundering if there is a easy way to draw the graphs on the calculator so I can find the intersections?

For example how do I plot:

25a + 10b <= 600

40a + 50b <= 2100

a <= 20

b <= 35

Would it not be easier to put the equations into a matrix and thence put it into reduced row-echelon form in order to solve the system? You can use the Rref command on the Prizm for that. If you still want to use the graphical method, you can specify the two equations to graph, but you'll have to convert them to the form y=f(x) first. In this case:

Let x=a and y=b.

25x+10y <= 600

y <= (600-25x)/10

40x + 50y <= 2100

y <= (2100-40x)/50

**KermMartian wrote:**

Would it not be easier to put the equations into a matrix and thence put it into reduced row-echelon form in order to solve the system? You can use the Rref command on the Prizm for that. If you still want to use the graphical method, you can specify the two equations to graph, but you'll have to convert them to the form y=f(x) first. In this case:

Let x=a and y=b.

25x+10y <= 600

y <= (600-25x)/10

40x + 50y <= 2100

y <= (2100-40x)/50

Thanks for the reply!

Would that just give me the awnser 17600?

I would like it to show me where my intercepts are. (0,0), (0,35), (35/4,35), (180/17,570/17) and so on.

But convert them to y=f(x) might be the easiest?
If he wants to put the equations in the graph, you don't need to solve for y first, because the Prizm can draw graphs for x (just select the TYPE with F3 on the graph expression list). Of course, if one needs to write down the steps one's doing on the calculator on the answer sheet (as exams in my country require), then one will still have to solve for y because the lowest-denominator is, that graphic calcs can only draw expressions in the form y=something. I have always solved linear programming problems by using the graph function.

To know the intercepts you'll need to use the intercept function (press F5 twice when seeing the graphs), and use it multiple times to know each intersection.

What is the objective function you're optimizing? Are a and b both implicitly bounded > 0?

In 2-dimensions, you pretty much just want to graph your constraints and check the corners.

If you write a program to draw your objective function as a gradient, and clip display to the inside of the constraints it should be really obvious which one you want.

Is it possible to rezise the grid behind the graphs so it's 1x1 instead of 10x10?

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