3.4: Linear Programming Intro: Oftentimes we want to optimize a situation - this means

to: find a maximum value (such as maximizing profits) find a minimum value (such as minimizing costs).

Linear Programming: Process of optimization on a linear function

Constraints: Boundaries (borders) on our function

Solution Set: called Feasible Region; shaded area of possibilities given our function and our constraints.

Linear Programming Example: Suppose a Cost

function is given by: C = 2x + 4y and we have the constraints:

x > 0, y > 0 and -x + 3y < 15, 2x + y < 12 (Use x & y intercepts to

graph) Find the feasible region. Solution: We already know

that x > 0 and y > 0 guarantee us an answer in the first quadrant. So let’s look at the other two.

-10 10

-10

10

x

y

Linear Programming Since our answer has to be

in the first quadrant, we know the feasible region is:

-10 10

-10

10

x

y

• Extension: Find the max value of C.

• Solution: The max value has to be the highest point of our feasible region.

This point is about: (3, 6)Since: C = 2x + 4yC = 2(3) + 4(6) = 6 + 24 = 30

Linear Programming Extenion: Find the min value

of C. Solution: The min value has

to be at the smallest point of our feasible region. This is clearly (0, 0)

C = 2x + 4y C = 2(0) + 4(0) = 0.

Note: If they exist, the MAX and MIN will always occur at one of the vertices of our bounded region.

-10 10

-10

10

x

y

What were the steps? Graph your system of inequalities (constraints)

Your function is NOT part of your system Shade carefully If it exists, the max value is the point at the highest

and farthest point (upper right) of your shaded polygon.

Otherwise, you have to plug in your vertices to see who the max is.

If it exists, the min value is at the lowest and smallest point (lower left) of your shaded polygon.

Otherwise, you have to plug in your vertices to see who the min is.

Make it happen Your turn: Given the function

C = 3x + 4y and the constraints: x > 0, y > 0, and x + y < 8:

A. Graph the system B. Find the max C. Find the min

-10 10

-10

10

x

y

The feasible region is the first quadrant right triangle.

Checking your answers out B. The max and min have to

occur at the vertices. Max: There are two

possibilities: C = 3x + 4y At (8,0), C = 3(8) + 4(0) = 24 At (0,8), C = 3(0) + 4(8) = 32 So our max is 32 and occurs

at (0, 8)

C. Min: C = 3(0) + 4(0) = 0 So our min is 0 and occurs at

(0, 0)

-10 10

-10

10

x

y

Unbounded Regions So far our feasible regions

have been some sort of polygon. That isn’t always the case.

Example: Suppose our function is C = 5x + 6y with the constraints: x > 0, y > 0, x + y > 5 and 3x + 4y > 18.

Graph/find the max and min. Solution: Because of x > 0, y

> 0, we know our feasible region is in the first quadrant, so let’s just worry about the other two constraints.

-10 10

-10

10

x

y

Unbounded Regions The feasible region is the

first quadrant orange area shown.

We can see clearly the feasible region continues forever upwards and therefore has no max value.

Notice we have vertices at (0, 5); (2, 3) and (6, 0). If we evaluate at all three

At (0, 5); C = 5(0) + 6(5) = 30 At (2, 3); C = 5(2) + 6(3) = 28 At (6, 0); C = 5(6) + 6(0) = 30 So our min occurs at (2, 3)

-10 10

-10

10

x

y