# How do we Maximize profit by Using a Search-Line?

📌 Example 1. A small business enterprise makes dresses and trousers. To make a dress requires ½ hour of cutting and 20 minutes of stitching. To make a trousers requires 15 minutes of cutting and ½ hour of stitching. The profit on a dress is USD40 and on a pair of trousers USD50. The business operates for a maximum of 8 hours per day.
Determine how many dresses and trousers should be made to maximize profit and what the maximum profit is.

✍ Solution:
Step 1: To solve the above problem we would have to translate the conditions or constraints from a verbal to a symbolic form. We first introduce our variables.

Let x be the number of dresses and y the number of trousers.

Step 2: Next we express the constraints as a system of inequalities.

x≥0 and y≥0, x and y being whole numbers ie. x, y∈No

Cutting Time: ½xy≤8 ie. 2x+y≤32

Stitching Time: ⅓xy≤8 ie. 2x+3y≤48

Step 3: Graph the system of inequalities and shade in the region that satisfy the constraints. The shaded region is called the feasible region. Step 4: Write the profit in terms of the variables.
P=40x+50y

Since the objective (in this case) is to make a maximum profit, the profit equation is called the objective function.

Step 5: To determine the maximum profit and the number of items that will give the maximum profit we can use one of two methods:

5.1 Substitute each of the ordered pairs of the vertices of the feasible region, excluding (0;0), ie. (0;16), (16;0) and (12;8) into profit equation.

(i) P=40(0)+50(16)

=0+800
=800

(ii) P=40(16)+50(0)

=640+0
=640

(iii) P=40(12)+50(8)

=480+400
=880

The maximum profit is USD880,00 for x=12 and y=8.

5.2 Using the search line:

In the objective function, make y the subject of the formula:

ie. The gradient of this line is -⅘ or any multiple of this, eg. -8/10, -12/15, etc.

Draw any broken line with this gradient. For maximum profit this line, called the SEARCH LINE, must be moved parallel to the first line until it passes through the highest point in the feasible region. This point is A(12;8). We substitute only this ordered pair in the profit equation to obtain the maximum profit as indicated above.

Steps to be followed in solving a Linear Programming Problem
1. Define the variables if they are not already defined in the problem, ie. Let x be … and y be ….
2. Write down the constraints in terms of the variables.
3. Graph the constraints and shade the Feasible Region.
4. Write down the Objective Function in terms of the variables. Using the gradient of the objective function, draw a Search Line. Use this line to maximise or minimise the objective function.

Further Examples
📌 Ex2. A farmer has 80 hectares of his farm available for planting maize and cabbages. He must grow at least 10 hectares of maize and 20 hectares of cabbages to meet demands. He prefers to plant more cabbages than maize but his work force and equipment will only allow him to cultivate a maximum of three times the quantity of cabbages to that of maize.
2.1 Represent the information above as a system of inequalities.
2.2 Sketch a graph of these inequalities.
2.3 If the profit on maize is USD800 per ha and on cabbages USD500 per ha, how should the farmer plant the two crops to make a maximum profit and what is this profit.
✍ Solution:
2.1 Let x be the number of hectares of maize and y the number of hectares of cabbages.
2.2 2.3 Draw a search line with a gradient of -80/50. The highest point in the feasible region through which the search line passes is the point (60;20).
The maximum profit is P=800(60)+500(20)=48000+10000=USD58000

📌 Ex3. A factory makes two types of beds, type A and type B. Each month x of type A and y of type B are produced.
The following constraints control monthly production:
(i) Not more than 50 beds of type A and 40 beds of type B can be made.
(ii) To show a profit at least 60 beds in all must be made.
(iii) The maximum number of beds that can be produced is 80.

3.1 The diagram shows the four constraints. Write down in terms of x and y the inequalities that represent these constraints.
3.3 If the objective function is given by the equation y=2x+P/150, where P is the monthly profit in USD, what is the profit per bed of the two types of bed.
3.4 How many of each type of bed must be produced per month to maximise profit? What is the maximum profit?
3.5 Explain how the production would be affected if the objective function was ✍ Solution:
3.1 0≤x≤50, 0≤y≤40, x+y≥260, x+y≤80.
3.2 3.3 P=300x+150y

Profit on type A is USD300 and on type B is USD150.

3.4 Draw a Search line with gradient -2. The highest point in the feasible region through which the search line passes is (50:30). ∴ 50 of type A and 30 of type B must be produced to maximise profit.
The maximum profit is P=300(50)+150(30)=19500 USD
3.5 Since the gradient of this objective function is -1, the upper limit of the search line
will coincide with the line x+y=80. In this case all the ordered pairs (x and y being whole numbers) satisfying the line within the feasible region will give the same maximum profit.
Let you read page Maximize each Objective Function to Get the Maximum!⛱️

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