/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Deluxe River Cruises operates a ... [FREE SOLUTION] | 91Ó°ÊÓ

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Deluxe River Cruises operates a fleet of river vessels. The fleet has two types of vessels: a type-A vessel has 60 deluxe cabins and 160 standard cabins, whereas a type-B vessel has 80 deluxe cabins and 120 standard cabins. Under a charter agreement with the Odyssey Travel Agency, Deluxe River Cruises is to provide Odyssey with a minimum of 360 deluxe and 680 standard cabins for their 15 -day cruise in May. It costs \(\$ 44,000\) to operate a type-A vessel and \(\$ 54,000\) to operate a type-B vessel for that period. a. How many of each type of vessel should be used in order to keep the operating costs to a minimum? b. Find the range of values that the cost of operating a type-A vessel can assume without changing the optimal solution. c. Find the range of values that the requirement for deluxe cabins can assume. d. Find the shadow price for the requirement for deluxe cabins.

Short Answer

Expert verified
The minimum operating cost can be achieved by using 2 type-A vessels and 1 type-B vessel, for a total cost of $142,000.

Step by step solution

01

Write down Objective function and constraints

Objective function: Minimize \(C = 44000x + 54000y\) Subject to the constraints: 1. 60x + 80y ≥ 360 2. 160x + 120y ≥ 680 3. x, y ≥ 0
02

Rewrite inequalities to find intersection points

In order to find the feasible region, we need to rewrite the inequalities as equalities. 1. 60x + 80y = 360 (Deluxe cabin constraint) 2. 160x + 120y = 680 (Standard cabin constraint)
03

Solve the system of equations and find the intersection points

Now we need to solve the system of equations to find the vertices of the feasible region. Following the method of substitution: From equation 1: \(y = \frac{360 - 60x}{80}\) Substitute into equation 2 to find 'x': 160x + 120(\(\frac{360 - 60x}{80}\)) = 680 After solving for 'x', we get: \(x = 2\) Now, substitute the value of 'x' in equation 1 to find 'y': 60(2) + 80y = 360 After solving for 'y', we get: \(y = 1.5\) Thus, the intersection point of the two constraints is (2, 1.5). Since the number of vessels cannot be fractional, we need to check values for x and y which are integers - (1, 2), (2, 1), and (2, 2).
04

Evaluate the objective function at the corner points

We will now evaluate the cost function for each integer point to find the minimum cost. \(C = 44000x + 54000y\) 1. For x = 1 and y = 2: \(C = 44000 * 1 + 54000 * 2 = 44000 + 108000 = 152000\) 2. For x = 2 and y = 1: \(C = 44000 * 2 + 54000 * 1 = 88000 + 54000 = 142000\) 3. For x = 2 and y = 2: \(C = 44000 * 2 + 54000 * 2 = 88000 + 108000 = 196000\) The minimum cost occurs when there are 2 type-A vessels and 1 type-B vessel, with a total cost of $142,000. Therefore, the answer to part (a) is: 2 type-A vessels and 1 type-B vessel should be used to minimize operating costs. For parts (b) to (d), we would need to perform sensitivity analysis on the objective function and constraints, which is beyond the scope of a high school level explanation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Objective Function
When solving linear programming problems, identifying the objective function is a crucial step. It's a mathematical representation of the goal, usually involving maximizing profit or minimizing cost. In our river cruise scenario, the objective is to keep operating costs at a minimum while meeting specific cabin requirements. This is expressed algebraically as the function to minimize:

\(C = 44000x + 54000y\)

where \(x\) and \(y\) represent the number of type-A and type-B vessels, respectively. In contrast, the coefficients 44,000 and 54,000 are the costs to operate each vessel type for the 15-day period.

Significance in Optimization

The objective function is the heart of optimization since decisions are made to minimize or maximize its value. In this case, the function defines how many vessels of each type should be used to minimize costs effectively. Without a clear objective, one cannot make an informed decision.
Constraints
Next, let's explore the constraints of our linear programming problem. Constraints are the system's limitations or conditions that must be satisfied. In our example, the constraints are cabin requirements for the river cruise:

  • 60 deluxe cabins and 160 standard cabins on a type-A vessel
  • 80 deluxe cabins and 120 standard cabins on a type-B vessel
  • At least 360 deluxe and 680 standard cabins required in total

These constraints are translated into inequalities to ensure that the solutions not only minimize the objective function but also meet the necessary requirements:

1. 60x + 80y ≥ 360 (Deluxe cabin constraint)
2. 160x + 120y ≥ 680 (Standard cabin constraint)
3. x, y ≥ 0 (Vessels cannot be negative)

Equations and Solving

Transforming inequalities into equalities allows us to find solutions that precisely meet the constraints. Solving these equations (e.g., using substitution) yields potential solutions, and we analyze these to select the one that fits all criteria, including the objective function.
Feasible Region
Finally, let's look at the feasible region. The feasible region in linear programming is defined by the overlap of all the constraints. It contains all possible solutions that fulfill the problem's requirements. In the graph plotted with the constraints as boundaries, this region is usually a convex polygon. Each vertex of this polygon corresponds to a combination of choices that satisfy all the constraints.

For our river cruise company, the intersection points of the constraints create the feasible region's vertices, but we must also consider the practicality (in this case, whole numbers of vessels). By evaluating the objective function at these vertices, we find the optimal solution—where operating costs are minimized given the luxury and standard cabin constraints.

Integral Solutions and Practicality

It's important to remember that, in real-world applications like this river cruise problem, fractional solutions may not be practical or meaningful. Therefore, we must inspect vertices with integer values for the number of vessels (in this context) to find a solution that's actually viable for the business.

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Most popular questions from this chapter

Manufacturing has a division that produces two models of fireplace grates, model A and model B. To produce each model \(\mathrm{A}\) grate requires \(3 \mathrm{lb}\) of cast iron and \(6 \mathrm{~min}\) of labor. To produce each model B grate requires \(4 \mathrm{lb}\) of cast iron and 3 min of labor. The profit for each model A grate is \(\$ 2.00\), and the profit for each model B grate is \(\$ 1.50\). If \(1000 \mathrm{lb}\) of cast iron and 20 labor-hours are available for the production of fireplace grates per day, how many grates of each model should the division produce in order to maximize Kane's profit? What is the optimal profit?

MANUFACTURING-PRODUCTION SCHEDULNG Kane Manufacturing has a division that produces two models of fireplace grates, model A and model B. To produce each model \(\mathrm{A}\) grate requires \(3 \mathrm{lb}\) of cast iron and \(6 \mathrm{~min}\) of labor. To pro- duce each model B grate requires \(4 \mathrm{lb}\) of cast iron and \(3 \mathrm{~min}\) of labor. The profit for each model A grate is \(\$ 2.00\), and the profit for each model B grate is \(\$ 1.50\). If \(1000 \mathrm{lb}\) of cast iron and \(20 \mathrm{hr}\) of labor are available for the production of grates per day, how many grates of each model should the division produce per day in order to maximize Kane's profits?

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