Solve the linear programming problem. Minimize and maximize

Question

z=40x+50y

subject to

3x+y>=24

x+y>=12

x+3y>=18

x,y>=0

Answer 1

The "boundary conditions" are:

3x + y > 24

x + y > 12

x + 3y > 18

Solving these for y gives us:

y > -3x + 24

y > -x + 12

y > -1/3x + 6

Graphing these and also implementing x > 0 and y > 0 we get a graph that looks like this:

The maximum solution to this problem will be on the boundaries.

But, we need to find out what point A is. It is the intersection of y = -3x + 24 and y = -1/3x + 6. To find out what the values of x and y are at point A, we set those 2 equations equal to each other.

-3x + 24 = -1/3x + 6

-9/3x + 1/3x = -24 + 6

-8/3x = -18

x = (-18)(-3/8) = 27/4 = 6.75

Using the first equation (y = -3x + 24), we plug in x = 6.75 to get y = 15/4 = 3.75.

Now, we can create a table by using z = 40x + 50y

x y z

0 6 300

8 0 320

6.75 3.75 457.5

So the maximum value of z is 457.5 (occurring when x = 6.75 and y = 3.75

The minimum value of z is 0 (occurring at x = 0 and y = 0)