Problem 21 - Precalculus: Functions and Graphs

Question

Stocking a fish farm A fish farmer will purchase no more than 5000 young trout and bass from the hatchery and will feed them a special diet for the next year. The cost of food per fish will be $$ 0.50\( for trout and $$ 0.75 for bass, and the total cost is not to exceed \) \( 3000\). At the end of the year, a typical trout will weigh 3 pounds, and a bass will weigh 4 pounds. How many fish of each type should be stocked in the pond in order to maximize the total number of pounds of fish at the end of the year?

Step-by-Step Solution

Answer

Stock 1000 trout and 4000 bass for a maximum of 19,000 pounds of fish.

Step 1: Define Variables

Let \( x \) represent the number of trout and \( y \) represent the number of bass to be stocked. Our objective is to find values of \( x \) and \( y \) that maximize the total fish weight while satisfying given constraints.

Step 2: Set Up the Objective Function

The objective is to maximize the total weight of the fish. A typical trout weighs 3 pounds, and a bass weighs 4 pounds at the end of the year. Thus, the total weight \( W \) can be expressed by the equation: \( W = 3x + 4y \).

Step 3: Establish Constraints

There are two constraints to consider:1. The total number of fish: \( x + y \leq 5000 \).2. The cost of food: \( 0.5x + 0.75y \leq 3000 \).

Step 4: Reconstruct Constraints for Solution

These constraints can be rewritten as:1. \( x + y \leq 5000 \) (total fish constraint).2. \( 2x + 3y \leq 6000 \) (by multiplying the cost equation by 2 to clear decimals).Non-negativity constraints are \( x \geq 0 \) and \( y \geq 0 \).

Step 5: Graph the Constraints

Graph the lines of constraints to determine the feasible region:1. The line \( x + y = 5000 \).2. The line \( 2x + 3y = 6000 \).Intersections and boundary lines define the feasible region.

Step 6: Identify Corners of the Feasible Region

The feasible region is determined by solving for intersection points:1. \( x = 0 \) and solving the second constraint for \( y \), gives \( y = 2000 \).2. \( y = 0 \) and solving the second constraint for \( x \), gives \( x = 3000 \).3. Intersection of lines \( x + y = 5000 \) and \( 2x + 3y = 6000 \). Solving these simultaneously gives \( x = 1000 \) and \( y = 4000 \).

Step 7: Calculate Objective Function at Each Corner

Evaluate the objective function at each corner:1. For \( (0, 2000) \): Weight = \( 3(0) + 4(2000) = 8000 \) pounds.2. For \( (3000, 0) \): Weight = \( 3(3000) + 4(0) = 9000 \) pounds.3. For \( (1000, 4000) \): Weight = \( 3(1000) + 4(4000) = 19000 \) pounds.

Step 8: Determine Maximum Weight

The maximum weight is obtained at the corner (1000, 4000) with a weight of 19,000 pounds.

Key Concepts

Objective Function

In linear programming, the objective function is a mathematical expression we want to maximize or minimize. It's like the goal of our decision-making process. In the fish farm problem, the objective function is to maximize the total weight of the fish. Each pound is related to the number of trout and bass we add. The function is represented as \( W = 3x + 4y \), where \( x \) is the number of trout and \( y \) is the number of bass. So, the objective is to find the best numbers of \( x \) and \( y \) that increase our fish weight to the maximum possible amount by the end of the year.

Constraints

Constraints are conditions or limitations that we must satisfy in a linear programming problem. They help define what's feasible or allowable for our situation. In the case of our fish farm, we have multiple constraints:

• Total number of fish constraint: \( x + y \leq 5000 \).
• Food cost constraint: \( 0.5x + 0.75y \leq 3000 \). To simplify, we multiply everything by 2 to get \( 2x + 3y \leq 6000 \).
• Additionally, we have non-negativity constraints: \( x \geq 0 \) and \( y \geq 0 \).

These constraints are like rules our solution must follow to be valid in real life. They can influence what combinations of \( x \) and \( y \) we can choose.

Feasible Region

The feasible region is the collection of all possible solutions that meet the constraints. It is typically visualized as a shaded area on a graph, where each axis represents one of our decision variables, \( x \) and \( y \). Using our constraints, we draw these constraints on the graph as lines and see where they overlap.

• The first line is \( x + y = 5000 \).
• The second line is \( 2x + 3y = 6000 \).

The feasible region is the area where both these conditions, along with the non-negativity conditions, are satisfied. This region can be tricky to see, but it represents all the combinations of trout and bass we could choose from that do not break any rules. Within this area, we can find our best solution by looking at the corner points.

Optimization Problem

An optimization problem in linear programming is where we have an objective function, like maximizing fish weight, and we need to look at all the constraints to determine the best solution available. We are seeking the maximum or minimum value achievable by the objective function within the feasible region. For our fish farm, solving the optimization problem involves systematically evaluating the objective function at all corner points of the feasible region. These points are:

• (0, 2000), with a total weight of 8,000 pounds.
• (3000, 0), with a total weight of 9,000 pounds.
• (1000, 4000), with a total weight of 19,000 pounds.

Comparing these results, the point (1000, 4000) gives the maximum weight of 19,000 pounds, making it the optimal stocking decision for the fish farm. This strategy ensures we use our resources efficiently to achieve the best possible outcome.