Question

A clothing company has 100 yards of cloth and produces shirts (x units) and vests (y units). Shirts require 10 units and have profit value of $5, while vests require 4 units and have profit value of $4. What is the optimal production solution? What if the company decides to also put a “non-zero constraint” on all production? Must produce at least 3 shirts and 10 vests.

Answer 1

Answer:

Decision Variables:

Let, x = units of shirt

y = units of vest

Objective Function: Objective is to maximize the total profit gained by producing shirts and vests

The profit values of shirt and vest are $5 and $4 per unit respectively

Thus, the profit function is given as follows:

Total Profit = profit per shirt x no. of shirts + (profit per vest x no. of vest)

Max Z = $5x + $4y

Subject To:

Company has 100 yards of cloth and each shirt and vest requires 10 units and 4 units of cloth respectively

Total cloth required by x shirts and y vest is given as 10x + 4y

Since company cannot use more than 100 yards of available cloth, the constraint is written as follows:

10x + 4y <= 100

Consider production units cannot be in negative quantities, x , y >= 0

Plotting of constraints on graph:

Convert the inequalities in the equality equation and find the X-axis and Y-axis intercepts to plot the equation lines.

The feasible area is OAB, by extreme point method determine the profit at points O, A, and B

The profit is highest at point B (0, 25), where x = 0 and y = 25

Optimal solution is: No. of shirts = 0, Number of vest = 25 and Profit = $100

Part 2: Must produce at least 3 shirts and 10 vests

Two more equation will be added in the formulation,

x >= 3 and y >= 10

Plot line x = 3 and Y = 10 on the graph, and the feasible region for both the equations will be away from the origin.

The polygon EFG is the feasible area where all the constraints are satisfied.

Point E is intersection of 10x + 4y = 100 and x = 3 lines,

Substitute x = 3 in equation 10x + 4y = 100, we get, y = (100 – 30)/4 = 17.5

Coordinates of E are (3, 17.5)

Point G is intersection of 10x + 4y = 100 and y = 10 equation lines,

Substitute y = 10 in equation 10x + 4y = 100, we get, x = (100 – 40)/10 = 6

Coordinates of G are (6, 10)

Point F is intersection of x = 3 and y = 10,

Coordinates of F are (3, 10)

The feasible area is EFG, by extreme point method determine the profit at points E, F, and G

The profit is highest at point e (3, 17.5), where x = 3 and y = 17.5

Optimal solution is: No. of shirts = 3, Number of vest = 17.5 and Profit = $85