4.2 Solving Linear Programming Problems
The goal of linear programming is to optimize
(minimize or maximize) situations that have
constraints.
A linear programming problem in two variables
looks like:
Minimize: ax + by
Subject to:
cx + dy ≤ e
fx + gy ≥ h
The expression ax + by is called the objective
function.
The inequalities are the constraints.
There are four steps to solve a linear
programming problem:
1. Graph the feasible region, the region
bounded by the inequalitite.
2. Find the coordinates of the corners of the
feasible region.
3. Substitute each (x, y) pair into the objective
function and simplify.
4. Choose the optimum value and state your
result clearly.
Example: #22 Podunk Institute of Technology
offers two math courses: Finite Math and
Applied Calculus. Each section of Finite math
has 60 students, and each section of Applied
Calculus has 50. The department is allowed to
offer a total of up to 110 sections. Furthermore,
not more than 6000 students will want to take a
math course (no student will want to take two
math courses). Suppose the university makes
a profit of $100,000/ section of FM course and
$50,000/ section of AC (the profit is the
difference between what the students are
charged and what the professors are paid).
How many sections of each course should the
department offer to make the largest profit?
Let
x be # sections FM
y be # sections AC
Maximize: p = 100,000x + 50,000y
Subject to:
Sections:
x + y ≤ 110
Students:
60x + 50y ≤ 6000
Reality:
x ≤ 0, y ≤ 0
In Calculator:
Y1 = –x +110
Y2 = -1.2x + 120
Graph, shading above since y is less than in
both.
Coordinates
p = 100,000x + 50,000y
(0, 0)
0
(0, 110)
5,500,000
(50, 60)
8,000,000
(100,0)
10,000,000
The maximum profit is $10 million, the school
should offer 100 sections of Finite Mathematics
and no Applied Calculus.
This isn’t realistic, but that is what the numbers
show.
2. The Megabuck Hospital Corp is to build a
state subsidized nursing home catering to
homeless patients as well as high income
patients. State regulations require that every
subsidized nursing home must house a
minimum of 1000 homeless patients and no
more than 750 high income patients in order to
qualify for state subsidies. The overall capacity
of the hospital is 2100 patients. The board of
directors, under pressure from a neighborhood
group, insists that the number of homeless
patients should not exceed twice the number of
high income patients. Due to the state subsidy,
the hospital will make an average profit of
$10,000 per month for every homeless patient,
whereas the profit per high income patient is
estimated at $8,000. How many of each type of
patient should it house in order to maximize
profit?
Let
x be the number of homeless patients
y be the number of high income patients
Maximize: p = 10,000x + 8,000y
Subject to:
Required minimum of homeless patients: x ≥
1000
Required maximum of high income patients: y ≤
750
Capacity: x + y ≤ 2100
Pressure from a neighborhood group: x ≤ 2y or
y ≥ 0.5x
In calculator: Set Xmax = 1000
Y1 = 750
Shade above
Y2 = -X +2100 Shade above
Y3 = 0.5X
Shade below
The feasible region has four corners.
Coordinates
p = 10,000x + 8,000y
(0, 0)
0
(0, 750)
6,000,000
(1350, 750)
19,500,000
(1400, 700)
19,600,000
Megabuck Hospital should house 1400
homeless people and 700 high income people
for a maximum profit of $19,000,000 per month.