I found this classical LP example online, and formulate , solve it using Matlab.

**A calculator company produces a scientific calculator and a graphing calculator. Long-term projections indicate an expected demand of at least****100****scientific and****80****graphing calculators each day. Because of limitations on production capacity, no more than****200****scientific and****170****graphing calculators can be made daily. To satisfy a shipping contract, a total of at least****200****calculators much be shipped each day.***x*: number of scientific calculators produced

*y*: number of graphing calculators produced*R*= –2*x*+ 5*y*, subject to:

100 <*x*< 200

80 <*y*< 170

*y*> –*x*+ 200

**If each scientific calculator sold results in a**

**$2**

**loss, but each graphing calculator produces a**

**$5**

**profit, how many of each type should be made daily to maximize net profits?**

The question asks for the optimal number of calculators, so my variables will stand for that:

Since they can’t produce negative numbers of calculators, I have the two constraints, *x* > 0 and *y* > 0. But in this case, I can ignore these constraints, because I already have that *x* > 100 and *y* > 80. The exercise also gives maximums: *x* < 200 and *y* < 170. The minimum shipping requirement gives me *x* + *y* > 200; in other words, *y* > –*x* + 200. The revenue relation will be my optimization equation: *R* = –2*x* + 5*y*. So the entire system is:

The feasibility region graphs as:

When you test the corner points at (100, 170), (200, 170), (200, 80), (120, 80), and (100, 100), you should obtain the maximum value of *R* = 650 at (*x*, *y*) = (100, 170). That is, the solution is “100 scientific calculators and 170 graphing calculators”.

C

OK, here is the sketch of how a human could use pen and paper to solve it.

It turns out that it’s not that straightforward to convert it into Matlab LP solver, linprog().

Matlab LP solver:

Here is the formulation:

Note that constraints x>=100 and y>=80 screw things up, because the Matlab LP solver only takes Ax<=b , note “<=”.

So, we need a smart formulation:

because the problem is asking “max”, but matlab solver is for “min”, we need to inverse them.

f=[2 -5]

a =

[1 0

-1 0

0 1

0 -1

-1 -1]

b =[200 -100 170 -80 -200]

Now, we could fit into the Matlab LP solver by running

linprog(f,a,b)

Result is the same as pen and paper solution!

>> linprog(f,a,b)

Optimization terminated.

ans =

100.0000 170.0000