Problem 352 - Calculus Volume 1

Question

For the following exercises, set up, but do not evaluate, each optimization problem. You have a garden row of 20 watermelon plants that produce an average of 30 watermelons apiece. For any additional watermelon plants planted, the output per watermelon plant drops by one watermelon. How many extra watermelon plants should you plant?

Step-by-Step Solution

Answer

Optimize the total yield function \( Y(x) = (20 + x)(30 - x) \) for non-negative integer \( x \leq 30 \).

Step 1: Define the Variables

Let’s denote the initial number of plants as \( n_0 = 20 \). The initial average yield per plant is \( y_0 = 30 \) watermelons per plant. Let \( x \) represent the number of extra watermelon plants added.

Step 2: Determine the New Yield per Plant

According to the problem, for each additional plant \( x \), the yield decreases by 1 watermelon per plant. Therefore, the new yield per plant is \( y = 30 - x \).

Step 3: Define the Total Yield Function

The total yield from all the plants, including the extra ones, is the product of the total number of plants and the yield per plant. The total number of plants is \( 20 + x \) and the yield per plant is \( 30 - x \). Thus, the total yield function \( Y \) is: \[ Y(x) = (20 + x)(30 - x) \].

Step 4: Set up the Optimization Problem

We need to maximize the total yield \( Y(x) \). Therefore, we must set up the problem as finding the value of \( x \) that maximizes the function:\[ Y(x) = (20 + x)(30 - x) \].

Step 5: Identify Constraints

Since \( x \) represents the number of additional plants, it must be a non-negative integer. Additionally, the yield per plant \( 30 - x \) must be non-negative, so \( x \leq 30 \).

Key Concepts

Calculus

Calculus is a branch of mathematics that is essential for solving optimization problems. It helps us to find the maximum or minimum values of functions, which can represent physical quantities like yield or profit.
In this exercise, calculus is used to determine the number of extra watermelon plants that maximize the total yield. By expressing the total yield as a function, we can apply calculus techniques to analyze it.

• Derivatives are used to find the rate of change of a function.
• A critical point, where the derivative is zero, could be a local maximum or minimum.
• Second derivative test helps determine the nature of a critical point.

To set up an optimization problem, we need to differentiate the yield function. This provides us with a strategy to find where the yield is maximized by setting the derivative to zero.

Yield Maximization

Yield maximization refers to increasing the productivity of a system. In this problem, it means finding how many extra watermelon plants can be planted without decreasing average yield per plant too much.
To maximize yield, you have to balance the number of plants and the yield per plant. Adding more plants lowers the yield per plant, while fewer plants might not use space efficiently.
The formula used here is:\[ Y(x) = (20 + x)(30 - x) \]This shows how yield changes with added plants. Our goal is to figure out the optimal number of plants to achieve the highest total yield.

Mathematical Modeling

Mathematical modeling involves creating a mathematical representation of a real-world situation to analyze and solve problems.
Here, it helps us transform the garden problem into an optimization problem using equations.
Here's what you need to know about the model in this problem:

• Define initial conditions like the number of plants and yield per plant.
• Identify how the yield changes with additional plants.
• Formulate the relationship into an equation or function.

For our garden problem, the model is the yield function \(Y(x)\). This captures how total yield depends on the number of extra plants. Mathematical models simplify complex scenarios into understandable terms, allowing us to apply mathematical techniques.

Constraints in Optimization

Constraints are conditions that your solution must satisfy. In mathematical optimization, constraints limit the values that variables can take.
For this problem, there are two main constraints on the variable \(x\), representing the number of extra plants:

• Non-negativity: \(x\) must be a whole number equal to or greater than zero.
• Yield per plant: The yield cannot go below zero, meaning \(30 - x\geq 0\).

These constraints ensure our solution is realistic. Constraints help maintain the feasibility and applicability of the solution.