Problem 36 - Calculus for Biology and Medicine

Question

Three different species of insects are reared together in a laboratory cage. They are supplied with two different types of food each day. Each individual of species 1 consumes 3 units of food \(A\) and 5 units of food \(B\), each individual of species 2 consumes 2 units of food \(A\) and 3 units of food \(B\), and individual of species 3 consumes 1 unit of food \(A\) and 2 units of food \(B\). Each day, 500 units of food \(A\) and 900 units of food \(B\) are supplied. How many individuals of each species can be reared together? Is there more than one solution? What happens if we add 550 units of a third type of food, called \(C\), and each individual of species 1 consumes 2 units of food \(C\), each individual of species 2 consumes 4 units of food \(C\), and each individual of species 3 consumes 1 unit of food \(C ?\)

Step-by-Step Solution

Answer

Initial solution: infinite possibilities with two food types. Adding food C: one solution with 50 of species 1, 50 of species 2, and 250 of species 3.

Step 1: Define the Variables

Let's define the variables:- Let \( x \) be the number of individuals of species 1.- Let \( y \) be the number of individuals of species 2.- Let \( z \) be the number of individuals of species 3.

Step 2: Set Up the Equations for Food A and Food B

We know from the problem statement that:- Individuals of species 1 consume 3 units of food \( A \); species 2 consume 2 units of food \( A \); species 3 consume 1 unit of food \( A \). Altogether, they consume 500 units. The food \( A \) equation is: \[ 3x + 2y + z = 500 \]- Individuals of species 1 consume 5 units of food \( B \); species 2 consume 3 units of food \( B \); species 3 consume 2 units of food \( B \). Altogether, they consume 900 units. The food \( B \) equation is: \[ 5x + 3y + 2z = 900 \]

Step 3: Solve the Linear System for Food A and Food B

Let's solve the system of equations:1. \(3x + 2y + z = 500\)2. \(5x + 3y + 2z = 900\)Using elimination or substitution, we can eliminate one of the variables and solve for the others. Here are the steps:- Multiply equation 1 by 2 to align \(z\), giving us \(6x + 4y + 2z = 1000\).- Subtract equation 2 from this new equation: \[ (6x + 4y + 2z) - (5x + 3y + 2z) = 1000 - 900 \] \[ x + y = 100 \]Then substituting \( y = 100 - x \) in equation 1: \[ 3x + 2(100 - x) + z = 500 \] \[ 3x + 200 - 2x + z = 500 \] \[ x + z = 300 \]Substituting \( x + y = 100 \) in the same, we get: \[ z = 300 - x \] If we choose \( x = 100 \), \( y = 0 \), and \( z = 200 \) or other combination like \( x = 70\), \( y = 30\), \( z = 230 \), both satisfy equations.

Step 4: Introduce Third Type of Food C and Adjust Equations

Given that each individual consumes additional units of food \( C \), we can create the equation for food \( C \):- Each individual of species 1 consumes 2 units of food \( C \).- Each individual of species 2 consumes 4 units of food \( C \).- Each individual of species 3 consumes 1 unit of food \( C \).This provides the equation: \[ 2x + 4y + z = 550 \]To solve with the introduced constraints, we plug the solutions from Steps 3 and solve again.

Step 5: Solve with All Three Food Constraints

Substitute the food constraint equations we previously set:1. \( x + y = 100 \)2. \( x + z = 300 \)3. \( 2x + 4y + z = 550 \)Substitute \( z = 300 - x \) in the third equation: \[ 2x + 4y + (300 - x) = 550 \] \[ x + 4y = 250 \]Combining the two equations from the system (99x = 50 99), \[ x = 50\] \[ y = 50 \] \[ z = 250 \]

Step 6: Verify the Solution

Verify that all equations are satisfied:- For food \( A \): \[ 3x + 2y + z = 3(50) + 2(50) + 250 = 150 + 100 + 250 = 500 \]- For food \( B \): \[ 5x + 3y + 2z = 5(50) + 3(50) + 500 = 250 + 150 + 550 = 900 \]- For food \( C \): \[ 2x + 4y + z = 2(50) + 4(50) + 250 = 100 + 200 + 250 = 550 \]All constraints are satisfied; therefore, there is a valid solution to the modified problem.

Key Concepts

Systems of Equations

Systems of equations are a collection of two or more equations with the same set of variables. In solving the original exercise, we were dealing with three variables representing the number of individuals from each insect species. The primary goal here was to find values for these variables that satisfy all the given equations simultaneously.

In the context of the exercise:

• Each equation represents a different kind of constraint or resource that must be allocated – specifically food types A, B, and potentially C.
• By setting up a system of equations, we can model the problem mathematically and apply various methods to find solutions.

For this exercise:

• The food A equation was given by \( 3x + 2y + z = 500 \).
• The food B equation was given by \( 5x + 3y + 2z = 900 \).

One of the techniques used to solve such systems is elimination, where equations are manipulated to eliminate one of the variables, eventually finding the values for all variables. The uniqueness of the solution depends on the system's structure, which can yield no solution, one solution, or infinitely many solutions.

Constraint Optimization

Constraint optimization involves finding the optimal solution to a problem under given constraints. In our case, the constraints were the daily food supply limits for types A, B, and C.

This exercise might seem simple, but it forms a basis for more advanced problems where optimization has a higher stake, such as maximizing profits or minimizing costs subject to constraints.

The adjusted inclusion of food C added complexity to the problem. Here, each species' consumption was described by:

• Species 1 consuming 2 units of food C,
• Species 2 consuming 4 units,
• Species 3 consuming 1 unit.

The new constraint was represented as \( 2x + 4y + z = 550 \).

Solving this modified system required making sure all food constraints were satisfied. This type of problem-solving is crucial because in real-world situations, resources are usually limited, and optimal allocation is essential for success.

Mathematical Modeling

Mathematical modeling involves creating mathematical representations of real-world scenarios to understand and solve complex problems. In this exercise, we turned a practical problem—feeding insects in a lab—into a set of mathematical equations.

The process of developing these models:

• Starts by defining variables that symbolize real-world items.
• Then, writes equations based on given information.
• Finally, solves these equations to interpret practical outcomes.

The beauty of mathematical models is their versatility; they allow us to adjust conditions, such as adding new variables—as we did with food type C—and still gain insights or find new solutions.

Creating these models involves understanding both the problem's constraints and how different factors interact. This skill is widely used across fields like economics, engineering, and environmental science, making it one of the most powerful tools in problem-solving.