Math Is Fun Forum

safaulk

Member

Registered: 2013-11-16

Posts: 1

Linear Programming using Graphing Calculator

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. 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/xjq0i38dz8j6j … 1%20PM.jpg

Last edited by safaulk (2013-11-16 08:33:43)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,163

Re: Linear Programming using Graphing Calculator

hi safaulk

Welcome to the forum. 

I don't know what calculator you are using so it is hard to determine why you sometimes get an error.

My graph below shows what this problem should look like. (click on it to see it larger) I did it using 'Sketchpad' as follows:

(i)  For -3x + 2y  = 8    I chose x = 0, and found y = 4, so (4,0) is a point on this line.

Then I chose x = 2 and found that y = 7, so (2,7) is another point on the line.

I connected them with a straight line to form the 'border' for the first constraint.

(ii)  Similarly I found (6,0) and (7,8) are points on the line  -8x + y = -48

The inequalities require points below the first line and above the second, so the solution space is the quadrilateral formed by these lines and the axes.

As the inequalities include =, points on the borders are also allowed.

(iii) To create the P = 13x + 2y I need to choose a value for P.  At this stage it doesn't matter what P is, so to keep the calculations easy I chose P = 26

That gives (2,0) and (0,13) as points on the line 13x + 2y = 26.  This line is shown in red.  As the problem is to maximise P, I want to slide this line up and right to get larger values of P, but keeping parallel to my first red line.

This 'best' line will be the one where the red line just touches a corner of the solution space.  With linear constraints, this will always be the case, so I looked at the top right corner where the lines cross.  That's at (8,16).  So the maximum P is when x = 8 and y = 16.

The calculation at the side shows this to be 136.

An answer of 137 is very close, so perhaps you were doing it correctly all along and the difference is just due to trying to read the point accurately.  I would always check by substitution anyway as graphs are not perfectly accurate even when you use a calculator.

Check:  (8,16) lies on -3x + 2y  = 8  as -3 x 8 + 2 x 16 = -24 + 32 = 8

(8,16) lies on -8x + y = -48 as -8 x 8 + 16 = -64 + 16 = -48

So that confirms the 'best' point.

Now calculate P in this case .... 13 x 8 + 2 x 16 = 104 + 32 = 136.

I don't understand why the calculator has 78 below the correct answer of 136. 

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob