Problem 8 - A First Course in Mathematical Modeling

Question

Cow to Market-A cow currently weighs \(800 \mathrm{lb}\) and is gaining \(35 \mathrm{lb}\) per week. It costs \(\$ 6.50\) a week to maintain the cow. The market price today is \(\$ 0.95\) per pound but is falling \(\$ 0.01\) per day. Formulate a mathematical model and find the optimal period to keep the cow until it is sold to maximize profits.

Step-by-Step Solution

Answer

Keep the cow for approximately 6 weeks to maximize profits.

Step 1: Define Variables and Functions

Let's define the variables. Let \( w(t) \) be the weight of the cow after \( t \) weeks, with an initial weight of \( 800 \) lbs. The function for weight is: \[ w(t) = 800 + 35t \] Let \( p(t) \) be the market price per pound of the cow after \( t \) weeks. The price is falling at \( 0.01 \) per day, which is \( 0.07 \) per week, from \( 0.95 \) per pound, so: \[ p(t) = 0.95 - 0.07t \] Define the profit function \( P(t) \) as the total revenue after selling the cow minus the cost to maintain it: \[ P(t) = w(t) \times p(t) - 6.50t \]

Step 2: Substitute Functions in Profit Formula

Substitute the weight and the price functions into the profit formula: \[ P(t) = (800 + 35t)(0.95 - 0.07t) - 6.50t \] Expand the expression to make it easier to handle. \[ P(t) = 800 \times 0.95 + 35t \times 0.95 - 800 \times 0.07t - 35t \times 0.07t - 6.50t \] Simplify further: \[ P(t) = 760 + 33.25t - 56t - 2.45t^2 - 6.50t \] \[ P(t) = 760 - 29.25t - 2.45t^2 \]

Step 3: Find the Maximum of the Profit Function

The profit function \( P(t) = 760 - 29.25t - 2.45t^2 \) is a quadratic equation in standard form \( ax^2 + bx + c \). The maximum of a quadratic equation \( ax^2 + bx + c \) with \( a < 0 \) occurs at \( t = -\frac{b}{2a} \). Here, \( a = -2.45 \) and \( b = -29.25 \). Apply the formula: \[ t = -\frac{-29.25}{2 \times -2.45} \] Solve for \( t \): \[ t = \frac{29.25}{4.9} \approx 5.97 \] Thus, the optimal number of weeks to keep the cow is approximately 6 weeks.

Key Concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation of degree two. It typically takes the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( x \) represents the variable. In the context of optimization, we often encounter quadratic equations when dealing with problems that have maximum or minimum points. This is because the graph of a quadratic equation is a parabola, which has a summit or a trough depending on the sign of \( a \).

• If \( a > 0 \), the parabola opens upwards, and the equation has a minimum point.
• If \( a < 0 \), the parabola opens downwards, and the equation has a maximum point.

In the problem about the cow, the profit function \( P(t) = 760 - 29.25t - 2.45t^2 \) is a quadratic equation in which \( a = -2.45 \), \( b = -29.25 \), and \( c = 760 \). Since \( a < 0 \), the parabola formed is downward opening, indicating a maximum profit point is available. To find this point, the formula for the vertex of a parabola, \( t = -\frac{b}{2a} \), is used. Applying this formula helps in identifying the exact period, in weeks, when the cow should be sold to maximize profits.

Optimization Problems

Optimization problems seek to find the best solution, particularly the maximum or minimum value of a function. These problems are common in economics, engineering, and operations research. When optimizing, you either want to maximize something good, like profits or minimize something bad, like costs.
In the cow problem, the goal is to maximize profit by deciding when to sell the cow. To achieve this, we need to set up a profit function and then determine the point at which this function reaches its maximum value. This involves looking at how the cow's weight and market price change over time, both of which are critical variables affecting total profit.

• Total profit depends on the weight of the cow and the price per pound.
• Expenses, like the weekly maintenance cost, also play a role in the total profit calculation.
• Identifying the exact week to sell the cow involves solving the profit equation for its peak value.

The step-by-step approach involves formulating the profit as a quadratic equation and finding its maximum using the properties of quadratic equations as described earlier. This methodical approach ensures that you derive the highest possible profit given the constraints of the problem.

Variables and Functions

Variables and functions are foundational concepts in mathematical modeling. Variables are symbols used to represent quantities that can change, while functions describe the relationship between these variables. In real-world problems, understanding variables and their respective functions allows us to construct mathematical models to solve complex situations.

Defining Variables:

In the cow example, we had:

• \( w(t) \) representing the cow's weight in pounds after \( t \) weeks. This function was \( w(t) = 800 + 35t \).
• \( p(t) \) representing the market price per pound after \( t \) weeks, formulated as \( p(t) = 0.95 - 0.07t \).

Formulating Functions:

Once the variables are defined, functions articulate how these variables interact. The function for weight demonstrates a linear increase as weeks go by, while the price function illustrates a decrease over time. These relationships are central to constructing the overall profit function \( P(t) \), which combines changes in weight and price along with weekly costs to highlight how profits evolve over time.
This step of defining variables and constructing functions is crucial as it sets the stage for further calculations, such as those required in the optimization process.