Practice!

Sample Problem

Find the minimum value of z=2x-y where y≥0, subject to the following constraints

y-x≥5

y+3x≤-5

x≤-3

Maximum value

3

-10

  -8

-2

Minimum value

3

-10

-8

  does not exist

Solution

 

First, sketch the region corresponding to the system of constraints.

y≥0 is the area above y=0 (the x-axis).

y-x≥5 is the area to the left of the line y=x+5

 y+3x≤-5 is the area to the left of the line y=-3x-5 and x≤-3 is the area to the left of the line x=-3. The constraints from the region highlighted in blue stripes.

Next, find the vertices of the region

At (-3, 4): z=2(-3)-(4)=-10

At (-3, 2): z=2(-3)-(2)=-8

At (-5, 0): z=2(-5)-(0)=-10

Note that the point (x, 0) lies in the region for all values x≤-5. Substituting this point into the objection function, you get z=2(x)+0=2x where x can be infinitely small. You can obtain values of z that are as small as you want. So, there is no minimum value of z. (a minimum value does not exist)

However, there is a maximum value of z (we are limited to positive y-values and negative x-values based on the constraints…the values of z will only get smaller and smaller as we move farther into the region, so the maximum value of z must occur at one of the vertices.

The maximum value of z is therefore -8 and occurs at x=-3 and y=2.