In order to solve problems which require application of linear programming, a graphical approach, it is necessary to

A typical problem requiring the method of linear programming, a graphical approach, provides linear constraints and an objective function, which is to be either maximized or minimized. We will refer for graphing purposes to a graphing calculator. An example of this type of problem is the following:

Bob builds tool sheds. He uses 10 sheets of dry wall and 15 studs for a small shed and 15 sheets of dry wall and 45 studs for a large shed. He has available 60 sheets of dry wall and 135 studs. If Bob makes $390 profit on a small shed and $520 on a large shed, how many of each type of building should Bob build to maximize his profit?

Let

*s* represent the number of small sheds and

represent the number of large sheds. Then the constraints upon this problem are the following:

P(

*s*,

) = 390

*s* + 520

We solve 10

*s* + 15

= 60 and 15

*s* + 45

= 135 for

and enter these into the calculator under the Y= menu, shading below these lines to see the region defined by the constraints.

= (-2/3)

*s* + 4

= (-1/3)

*s* + 3

We

graph in the first

quadrant since both

*s* and

are greater than or equal to zero. The screens from the calculator are shown below.

We maximize the objective

function P(

*s*,

) = 390s + 520

by evaluating it at the vertices: (6,0), (3,2), and (0,3).

We have:

The final

equation shows us that Bob can maximize his profit at $2340 if he can sell 6 small sheds and not make any large sheds.

Notice that in order to maximize the objective function, the given information must be converted into inequalities, graphed, and the objective

function must be evaluated at the vertices of the polygonal region graphed.

Notice also that the minimum value would be obtained at the

vertex (0, 0) which we did not previously

evaluate since it would give a value of $0 for the profit.

**
Remember: Both the minimum and maximum values of the objective function are found at the vertices.
**