Solving systems of inequalities has an interesting application--it allows us to find the minimum and maximum values of quantities with multiple constraints.

First, assign a variable (
*x*
or
*y*
) to each quantity that is being
solved for. Write an equation for the quantity that is being maximized
or minimized (cost, profit, amount, etc.). This is that
maximization or minimization equation. Next, write each
constraint as an inequality. Number each inequality and graph the
system, numbering each line on the graph as its corresponding
inequality.

You should now have a shaded solution region with several "corners." Each corner is the intersection of two constraint inequalities. Find the coordinates of the corners by solving the systems of intersecting equations.

Plug the coordinates of the corners into the maximization/minimization equation. The coordinates that give the largest or smallest value for this equation (depending on what the problem is looking for) are the solution to the problem.

There are three quantities that we are often asked to maximize and minimize in linear programming problems. Revenue is the total amount of money taken in, cost is the total amount of money spent, and profit is the revenue minus the cost, or the total amount of money gained.

*Example*: Jimmy is baking cookies for a bake sale. He is making
chocolate chip and oatmeal raisin cookies. He gets 25 cents for each
chocolate chip cookie and 30 cents for each oatmeal raisin cookie. He
cannot make more than 500 cookies of each kind, and he cannot make more
than 800 cookies total. He must make at least one-third as many
chocolate chip cookies as oatmeal raisin cookies. How many of each kind
of cookie should he make to get the most money?

**Variables:**
*x*
= number of chocolate chip cookies (in
hundreds)

*y*
= number of oatmeal raisin cookies (in hundreds)

**Maximization equation:** Profit =
25*x* + 30*y*

**Constraints**:

*1.*
*x*≤5

*2.*
*y*≤5

*3.*
*x* + *y*≤8

*4.*
*x*≥
*y*

**Graph:**

Graph of Constraints

1and3:x= 5 ,x+y= 8 . (x,y) = (5, 3) .

2and3:y= 5 ,x+y= 8 . (x,y) = (3, 5) .

2and4:y= 5 ,x=y. (x,y) = (, 5) .

(5, 3)âá’Profit = 25(5) + 30(3) = 215 .

(3, 5)âá’Profit = 25(3) + 30(5) = 225 .

(, 5)âá’Profit = 25() + 30(5) = 191.67 .

Thus, the

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