 Suggested languages for you:

Europe

Problem 11

# A company manufactures two products, $$\mathrm{A}$$ and $$\mathrm{B}$$, on machines $$\mathrm{I}$$ and $$\mathrm{II}$$. The company will realize a profit of $$\ 3 /$$ unit of product $$A$$ and a profit of $$\ 4 /$$ unit of product $$B$$. Manufacturing 1 unit of product A requires 6 min on machine I and 5 min on machine II. Manufacturing 1 unit of product $$\mathrm{B}$$ requires 9 min on machine I and 4 min on machine II. There are 5 hr of time available on machine I and $$3 \mathrm{hr}$$ of time available on machine II in each work shift. a. How many units of each product should be produced in each shift to maximize the company's profit? b. Find the range of values that the contribution to the profit of 1 unit of product A can assume without changing the optimal solution. c. Find the range of values that the resource associated with the time constraint on machine I can assume. d. Find the shadow price for the resource associated with the time constraint on machine $$\mathrm{I}$$,

Expert verified
In summary, to maximize profit, the company should produce 20 units of product A and 20 units of product B per shift, with a maximum profit of $140. The range of values for product A's profit is $$[4, +\infty)$$, and the range for the resource constraint of machine I is $$[180, 360]$$. The shadow price for machine I's time constraint is$5.
See the step by step solution

## Step 1: Let $$x_1$$ be the number of units of product A produced and $$x_2$$ be the number of units of product B produced. We have the objective function to maximize total profit: $P = 3x_1 + 4x_2$ Subject to the following constraints: $6x_1 + 9x_2 \le 300$ (Time constraint on machine I, since 5 hours = 300 minutes) $5x_1 + 4x_2 \le 180$ (Time constraint on machine II, since 3 hours = 180 minutes) and $x_1, x_2 \ge 0$ (Non-negativity constraints) #Step 2: Graph the Feasible Region#

Plot the constraints on a coordinate plane with $$x_1$$ on the x-axis and $$x_2$$ on the y-axis to find the feasible region. The first constraint yields $$x_2 \le \frac{300 - 6x_1}{9}$$, while the second constraint yields $$x_2 \le \frac{180 - 5x_1}{4}$$. The feasible region is determined by these inequalities and the non-negativity constraints. #Step 3: Identify the Corner Points of the Feasible Region#

## Step 2: Four corner points are formed by the intersection of the constraints: (0, 0), (0, 20), (20, 20), and (30, 0). #Step 4: Evaluate Objective Function at Each Corner Point#

Find the objective function value for each corner point: 1. At (0, 0): $$P = 3(0) + 4(0) = 0$$ 2. At (0, 20): $$P = 3(0) + 4(20) = 80$$ 3. At (20, 20): $$P = 3(20) + 4(20) = 140$$ 4. At (30, 0): $$P = 3(30) + 4(0) = 90$$ The objective function has its maximum value of 140 at the point (20, 20). #Step 5: Answer Part a#

## Step 3: The company should produce 20 units of product A and 20 units of product B in each shift to maximize the profit. #Step 6: Find the Sensitivity Range for Product A Profit#

To find the range of values for the profit of product A, we must calculate the slope coefficient of product A's objective function so that the optimal solution at (20, 20) remains the same. The slope for product A's objective function is $$\frac{-3}{4}$$. We need to find the minimum and maximum slopes of the isoprofit lines tangent to the feasible region corners adjacent to (20, 20). The minimum slope is found at the origin (0, 0). The maximum slope is found at the intersection of machine II's time constraint and the non-negativity constraint (30, 0). Hence, the minimum value for the slope without changing the optimal solution is $$\frac{-3}{4}$$, and the maximum slope is $$\frac{-5}{4}$$. Consequently, the range of values for the profit of product A is $$[4 ,+\infty)$$. #Step 7: Find the Sensitivity Range for Constraint on Machine I#

## Step 4: The sensitivity range for the constraint on machine I can be determined by examining the change in the constraint coefficient that does not change the optimal corner point. It can be found by calculating the slack values in the linear constraints. Slack of the first constraint at (20, 20): $$300 - 6(20) - 9(20) = -60$$, which indicates overutilization of machine I. The range for the resource constraint of machine I is $$[180 , 360]$$. #Step 8: Find the Shadow Price for Machine I#

Shadow price, also known as dual variable, for machine I is the amount the objective function's value will change per additional minute of availability of machine I. We find it through marginal increase in profit. At the intersection of the isoprofit line and machine I's constraint line, we have: $$\frac{4}{3} (300 - 6x_1 - 9x_2) = 180 - 5x_1 - 4x_2$$ Solve for $$x_1, x_2$$: $$x_1 = 15;\ x_2 = 25$$ Marginal increase in profit = $$3 \cdot 15 + 4 \cdot 25 - 3 \cdot 20 - 4 \cdot 20 = 5$$ The shadow price for the resource associated with the time constraint on machine I is \$5.

We value your feedback to improve our textbook solutions.

## Access millions of textbook solutions in one place

• Access over 3 million high quality textbook solutions
• Access our popular flashcard, quiz, mock-exam and notes features ## Join over 22 million students in learning with our Vaia App

The first learning app that truly has everything you need to ace your exams in one place.

• Flashcards & Quizzes
• AI Study Assistant
• Smart Note-Taking
• Mock-Exams
• Study Planner 