Question

A manufacturer has three machines A, B, and C. Machines A and B can be operated for at most 12 hours whereas machine C must be operated for at least 5 hours a day. These three machines produce only two items X and Y. Product X requires 1 hour from machine A, two hours from machine B and 1 hour from machine C. Product Y requires 2 hours from machine A, one hour from machine B and 1.25 hours from machine C. The manufacturer is required to sell at least 1 item from product X and 2 items from product Y. The manufacturer makes a profit of $600 and $400 on items X and Y, respectively. Using linear programming with a graphical approach, how many items should the manufacturer produce to maximize the profit? What will be the maximum profit? ______________.

Please Solve As soon as
Solve quickly I get you UPVOTE directly
Thank's
Abdul-Rahim Taysir

Answer 1

Based on the data given, we can write the following table:

Profit from the production of X= Profit per unit*Number of units

=$600*X

=600X

Profit from the production of Y= Profit per unit*Number of units

=$400*Y

=400Y

Let Z be the profit function. It can be expressed as:
Z=600X+400Y
x≥0
y≥0

The production of both the units will be zero or greater than zero.
X+2Y≤12 (Limitation of A)
2X+Y≤12 (Limitation of B)
X+5/4Y≥5 (Limitation of C)
BY PLOTTING GRAPHICAL POINTS, we get the following graph:

The common are of the graph where the profit will be maximised is given below in the blue area alongwith the points:

The different points along with their level of profits are:

A[0,6]=600*0+400*6=2400
B[0,4]=600*0+400*4=1600
C[4,4]=600*4+400*4=4000
D[6,0]=600*6+400*0=3600
E[5,0]=600*5+400*0=3000
so max value is at , X=4;Y=4

So, Number of units produced of X=4

Number of units produced of Y=4

The maximum profit is $4,000.