Problem 41 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The water-supply manager for a Midwest city needs to supply the city with at least 10 million gal of potable (drinkable) water per day. The supply may be drawn from the local reservoir or from a pipeline to an adjacent town. The local reservoir has a maximum daily yield of 5 million gal of potable water, and the pipeline has a maximum daily yield of 10 million gallons. By contract, the pipeline is required to supply a minimum of 6 million gallons/day. If the cost for 1 million gallons of reservoir water is $$\$ 300$$ and that for pipeline water is $$\$ 500$$, how much water should the manager get from each source to minimize daily water costs for the city? What is the minimum daily cost?

Step-by-Step Solution

Answer

The manager should get either 4 million gallons from the reservoir and 6 million gallons from the pipeline or 5 million gallons from the reservoir and 6 million gallons from the pipeline to minimize the daily water costs. The minimum daily cost will be $$\$ 4200$$.

Step 1: Define the decision variables

Let x be the amount of water (in millions of gallons) taken from the reservoir and y be the amount of water (in millions of gallons) taken from the pipeline.

Step 2: Formulate the objective function

We want to minimize the daily cost. The cost for 1 million gallons of reservoir water is $$\$ 300$$ and that for pipeline water is $$\$ 500$$. Therefore, the objective function for the total cost can be written as: Minimize: \(Z = 300x + 500y\)

Step 3: Write the constraints

According to the problem description, we have the following constraints: 1. The city needs at least 10 million gallons of water per day: \( x + y \geq 10 \) 2. The maximum daily yield from the reservoir is 5 million gallons: \( x \leq 5 \) 3. The maximum daily yield from the pipeline is 10 million gallons: \( y \leq 10 \) 4. The minimum daily yield from the pipeline is 6 million gallons: \( y \geq 6 \) 5. The quantities of water taken from both sources must be non-negative: \( x \geq 0, y \geq 0 \)

Step 4: Graph the feasible region

Plot the constraints on a graph to find the feasible region: 1. \( x + y \geq 10 \) is above the line \( x + y = 10 \). 2. \( x \leq 5 \) is to the left of the line \( x = 5 \). 3. \( y \leq 10 \) is below the line \( y = 10 \). 4. \( y \geq 6 \) is above the line \( y = 6 \). After plotting these lines, we see that the feasible region is a quadrilateral with vertices at points A(4, 6), B(5, 6), C(5, 10), and D(0, 10).

Step 5: Determine the optimal solution

Evaluate the objective function at each vertex to find which one gives the minimum cost: 1. At point A(4, 6): \(Z = 300(4) + 500(6) = 4200\) 2. At point B(5, 6): \(Z = 300(5) + 500(6) = 4200\) 3. At point C(5, 10): \(Z = 300(5) + 500(10) = 6300\) 4. At point D(0, 10): \(Z = 300(0) + 500(10) = 5000\) The minimum cost of \(Z = 4200\) occurs at points A(4, 6) and B(5, 6), meaning the manager can choose to get 4 million gallons from the reservoir and 6 million gallons from the pipeline or 5 million gallons from the reservoir and 6 million gallons from the pipeline to achieve the minimum daily cost of $$\$ 4200$$.

Key Concepts

Optimization Problems

Optimization is the process of making a system or design as effective or functional as possible. In the realm of mathematics, optimization refers to finding the 'best' solution to a problem within a defined set of constraints and conditions. Linear programming is a powerful mathematical method used to solve optimization problems, where the relationship between variables is linear. An optimization problem involves an objective function, constraints, and a feasible region. The goal is to maximize or minimize the objective function, like costs or profits, while adhering to the given constraints.

Objective Function

The objective function is the heart of every optimization problem in linear programming. It's the mathematical expression that defines what needs to be maximized or minimized. In the case of the Midwest city's water supply problem, the objective function represents the total cost, which is calculated as the sum of the costs of water from each source. The function is formulated based on unit costs and quantities, seen as

Minimize: \(Z = 300x + 500y\)

Here, \(x\) and \(y\) represent the quantities obtained from different sources, and multiplication with their respective costs gives us the total expenditure to be minimized.

Feasible Region

The feasible region is a pivotal concept in linear programming as it represents all possible solutions to the problem that meet the constraints. It is where all requirements, such as resource limitations and minimum or maximum demands, overlap. This region is graphically depicted and often bounded by lines or planes in higher dimensions. In our Midwest city example, we identify the feasible region on a graph where the axes represent the amount of water from each source. The lines drawn according to constraints enclose the region where the solution to the problem lies.

Constraint Formulation

Constraints are the rules that must be followed within an optimization problem. These often embody the limitations or requirements, like minimum or maximum resource capacities. In formulating constraints, inequalities are commonly used to represent these limitations. For example, the constraint that the pipeline must supply at least 6 million gallons of water is represented as \(y \geq 6\). Each constraint contributes to the shaping of the feasible region, setting the stage for where the optimal solution can be found.

Cost Minimization

Cost minimization is a frequent objective in optimization problems to reduce expenses under a given set of constraints. It's not just about finding the lowest cost but doing so within the realm of feasible solutions. In the case of the water-supply manager, cost minimization is pivotal in efficient municipal management. By combining the objective function with a graphical understanding of the feasible region and applying the given constraints, the manager can easily determine the combination of water sources that leads to the lowest cost without compromising on the city's water needs.