A linear programming model consists of two basic pieces; an objective function which you want to maximize or minimize, and a series of constraints. The constraints are written as linear inequalities while the objective function is a linear equation.

Let A be the number of type A built, and B the number of type B.

In this problem you want to maximize profit. **The objective function
is** `P=76A+48B` .(The objective function is often written
as C= some linear function. I used P to indicate profit)(The total profit is 76
times the number of type A plus 48 times the number of type B). This is a
straight line, and thus has no maximum. However, you are bound by certain
restraints. For instance, you might have limited access to parts, limited
number of employees, etc...

In this problem we are given a number of constraints -- restrictions that limit the number of bikes that can be made:

`A>=0,B>=0` These are the "natural" constraints. We disregard any negative solutions as they are meaningless in the real world. You might have different constraints -- if you are told you must produce at least 10 type A you would have `A>=10`

`A<=100,B<=150` These constraints are given in the problem as the maximum number of bikes that can be built.

`62A+42B<=6920` The number of hours spent making type A bikes is 62A since each bike takes 62 hrs. Similarly, the amount of time for B bikes is 42B. There is a total of 6920 hrs for assembly.

`78A+48B<=8280` These constraints reflect the restriction requiring a certain number of hours for packaging.

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The objective function is `P=76A+48B`

The constraints are:

`0<=A<=100,0<=B<=150`

`62A+42B<=6920`

`78A+48B<=8280`

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Now you graph all of the constraints simultaneously on a coordinate plane. Since these are inequalities, you will end up with a region that includes every point that is a solution to every constraint. This is called the feasible region.

There is a theorem that indicates that any maximum or minimum must occur at the boundaries of the feasible region. Thus you need only look at the corners. (There are some assumptions being made about the corners only having whole number coordinates and the feasible region being closed and convex, but this will turn out to be the case here)

So graph the constraints and locate the corners -- where the lines cross.

**Whether looking at the graph or solving the systems, we get the
following corners, starting at the origin and moving clockwise:**

**(0,0),(0,150),(10,150),(52,88),(100,10),(100,0)**

To find the solution, we plug each (A,B) point into the objective function and pick the maximum.

P(0,0)=0

P(0,150)=7200

P(10,150)=7960

P(52,88)=8176

P(100,10)=8080

P(100,0)=7600

**Thus the most profit can be made when you produce 52 type A, and 88
type B**

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