Linear Programming Solved Problems — How do we Maximize the Profit of each Problem?

📌 Solved-Problem 1. A merchant plans to sell two types of personal computers — a desktop model and a portable model that will cost $25000 and $40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than $7 million and if his profit on the desktop model is $4500 and on portable model is $5000.
✍ Solution:
Let the merchant stock x desktop models and y portable models. Therefore, x≥0 and y≥0.
The cost of a desktop model is $25000 and of a portable model is $4000. However, the merchant can invest a maximum of $7 million.

∴25000x+40000y≤7000000 … (÷5000)
5x+8y≤1400

The monthly demand of computers will not exceed 250 units.
x+y≤250

The profit on a desktop model is $4500 and the profit on a portable model is $5000. Total profit, Z=4500x+5000y.
Thus, the mathematical formulation of the given problem is maximizing Z=4500x+5000y … (1)
subject to the constraints,
5x+8y≤1400 … (2)
x+y≤250 … (3)
x,y≥0 … (4)

The feasible region determined by the system of constraints is as follows.
the system of constraints

The corner points are A(250, 0), B(200, 50), and C(0, 175). The values of Z at these corner points are as follows.

Corner point Z=4500x+5000y
A(250, 0) 1125000
B(200, 50) 1150000 → Maximum
C(0, 175) 875000

The maximum value of Z is 1150000 at (200, 50). Thus, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of $1150000.

📌 Solved-Problem 2. Sarah makes bracelets and necklaces to sell at a craft store. Each bracelet makes a profit of $7, takes 1 hour to assemble, and costs $2 for materials. Each necklace makes a profit of $12, takes 2 hour to assemble, and costs $3 for materials.

Sarah has 48 hours available to assemble bracelets and necklaces. If she has $78 available to pay for materials, how many bracelets and necklaces should she make to maximize her profit?
✍ Solution:
To formulate this as a linear programming problem:
1. Identify the variables.
2. Write the objective function.
3. Write the constraints.

1. Let x=the number of bracelets Sarah makes
Let y=the number of necklaces Sarah makes
2. Express the profit as a function of x and y.

p=7x+12y ⟷[Function to be maximized]

3. Express the constraints as inequalities.
Cost of materials: 2x+3y≤78.
Time limitation: x+2y≤48.

Since Sarah cannot make a negative number of bracelets or necklaces, x≥0 and y≥0 must also hold.

Maximize p=7x+12y subject to the constraints

2x+3y≤78, x+2y≤48, x≥0, and y≥0.

constraints

Sarah should make 12 bracelets and 18 necklaces for a maximum profit of $300.

📌 Solved-Problem 3. We want to solve the following LP problem for Failsafe Computers using the corner-point method:
Maximize profit =$9X1+$7X2

2X1+1X2≤40
X1+3X2≤30

feasible region e

FIGURE SP3 ■ Failsafe Computers‘ Feasible Region

✍ Solution:
Figure SP3 illustrates these constraints:

Corner-point a: (X1=0, X2=0) Profit =0
Corner-point b: (X1=0, X2=10) Profit: 9(0)+7(10)=$70
Corner-point d: (X1=20, X2=0) Profit: 9(20)+7(0)=$180

Corner-point c is obtained by solving equations 2X1+1X2=40 and X1+3X2=30 simultaneously. Multiply the second equation by -2 and add it to the first.

2X1+1X2=40
-2X1-6X2=-60
_______________
-5X2=-20

Thus X2=4.
X1+3(4)=30 or X1+12=30 or X1=18.
Corner-point c: (X1=18, X2=4) Profit=9(l8)+7(4)=$190.

Hence the optimal solution is
(X1=18, X2=4) Profit=$190.

📌 SP4. A local clinic wants to produce a guide to healthy living. The clinic intends to produce the guide in two formats: a short video and a printed book. The clinic needs to decide how many of each format to produce for sale. Estimates show that no more than 10,000 copies of both items together will be sold. At least 4000 copies of the video and at least 2000 copies of the book could be sold, although sales of the book are not expected to exceed 4000 copies. Let x be the number of videos sold, and y the number of printed books sold.
a) Write down the constraint inequalities that can be deduced from the given information.
b) Represent these inequalities graphically and indicate the feasible region clearly.
c) The clinic is seeking to maximise the income, I, earned from the sales of the two products. Each video will sell for $50 and each book for $30. Write down the objective function for the income.
d) What maximum income will be generated by the two guides?
✍ Solution:
a) Estimates show that no more than 10,000 copies of both items together will be sold:

x+y≤10,000

At least 4000 copies of the video could be sold:
x≥4000

At least 2000 copies of the book could be sold:
y≥2000

Sales of the book are not expected to exceed 4000 copies:
y≤4000

b)
graph b

c)
I=($50)×(videos sold)+($30)×(printed books sold)
I=50x+30y

d) The vertices of the feasible region are as follows:
Point A: (4000; 2000)
Point B: (4000; 4000)
Point C: (6000; 4000)
Point D: (8000; 2000)

The cost at each vertex is as follows:
Point A: I=50(4000)+30(2000)=260,000
Point B: I=50(4000)+30(4000)=320,000
Point C: I=50(6000)+30(4000)=420,000
Point D: I=50(8000)+30(2000)=460,000

The maximum profit of $460,000 can be made if 8000 videos and 2000 books are sold.

📌 SP5. A group of students plan to sell at hamburgers and y chicken burgers at a rugby match. They have meat for at most 300 hamburgers and at most 400 chicken burgers. Each burger of both types is sold in a packet. There are 500 packets available. The demand is likely to be such that the number of chicken burgers sold is at least half the number of hamburgers sold.
a) Write the constraint inequalities and draw a graph of the feasible region.
b) A profit of $3 is made on each hamburger sold and $2 on each chicken burger sold. Write the equation which represents the total profit P in terms of x and y.
c) The objective is to maximise profit. How many of each type of burger should be sold?
✍ Solution:
a) They have meat for at most 300 hamburgers:

x≤300

They have meat for at most 400 chicken burgers:
y≤400

The demand is likely to be such that the number of chicken burgers sold is at least half the number of hamburgers sold:
y≥(0.5)x

Each burger of both types is sold in a packet. There are 500 packets available:
x+y≤500

hamburger profit sale

b)
P=(profit per hamburger)×(number of hamburgers sold)
+(profit per chicken burger) × (number of chicken burgers sold)
P=3x+2y

c) The vertices of the feasible region are as follows:
Point A: (0; 0)
Point B: (300; 150)
Point C: (300; 200)
Point D: (100; 400)
Point E: (0; 400)

The cost at each vertex is as follows:
Point A: P=3(0)+2(0)=0
Point B: P=3(300)+2(150)=1200
Point B: P=3(300)+2(200)=1300
Point D: P=3(100)+2(400)=1100
Point E: P=3(0)+2(400)=800

To maximize the objective function select Point C.
The maximum possible profit of $1300 can be made if 300 hamburgers and 200 chicken burgers are sold.

📌 SP6. A calculator company produces a scientific calculator and a graphing calculator. Long-term projections indicate an expected demand of at least 100 scientific and 80 graphing calculators each day. Because of limitations on production capacity, no more than 200 scientific and 170 graphing calculators can be made daily. To satisfy a shipping contract, a total of at least 200 calculators much be shipped each day.
If each scientific calculator sold results in a $2 loss, but each graphing calculator produces a $5 profit, how many of each type should be made daily to maximize net profits?
✍ Solution:
The question asks for the optimal number of calculators, so my variables will stand for that:

x: number of scientific calculators produced
y: number of graphing calculators produced

Since they can’t produce negative numbers of calculators, I have the two constraints, x≥0 and y≥0. But in this case, I can ignore these constraints, because I already have that x≥100 and y≥80. The exercise also gives maximums: x≤200 and y≤170. The minimum shipping requirement gives me x+y≥200; in other words, y≥–x+200. The profit relation will be my optimization equation: P=–2x+5y. So the entire system is:
P=–2x+5y, subject to:
100≤x≤200
80≤y≤170
y≥–x+200

The feasibility region graphs as:
linear programming solved problem

When you test the corner points at (100, 170), (200, 170), (200, 80), (120, 80), and (100, 100), you should obtain the maximum value of P=650 at (x, y)=(100, 170).
💪 How do we Maximize profit by Using a Search-Line? 💎

RELATED POSTs