Problem 70 - Algebra and Trigonometry

Question

Agriculture At a certain vineyard it is found that each grape vine produces about 10 pounds of grapes in a season when about 700 vines are planted per acre. For each additional vine that is planted, the production of each vine decreases by about 1 percent. So the number of pounds of grapes produced per acre is modeled by $$ A(n)=(700+n)(10-0.01 n) $$ where \(n\) is the number of additional vines planted. Find the number of vines that should be planted to maximize grape production.

Step-by-Step Solution

Answer

Plant 850 vines per acre to maximize grape production.

Step 1: Understand the problem

We need to find the number of additional vines, denoted by \( n \), that should be planted to maximize the grape production per acre.

Step 2: Define the function

The function to maximize is given by \[ A(n) = (700 + n)(10 - 0.01n) \]. This represents the total production of grapes per acre based on the number of initially planted vines plus additional vines \( n \) and the reduced production per vine.

Step 3: Expand the function

Expand the function \( A(n) \) to make it easier to find the maximum value. \[ A(n) = (700 + n)(10 - 0.01n) = 7000 + 10n - 7n - 0.01n^2 = -0.01n^2 + 3n + 7000 \]

Step 4: Find the derivative

To find the number of additional vines \( n \) that maximizes \( A(n) \), first compute the derivative of \( A(n) \) with respect to \( n \): \[ A'(n) = \frac{d}{dn}(-0.01n^2 + 3n + 7000) = -0.02n + 3 \]

Step 5: Find the critical points

Set \( A'(n) = 0 \) to find the critical points:\[ -0.02n + 3 = 0 \]Solving for \( n \): \[ -0.02n = -3 \]\[ n = \frac{3}{0.02} = 150 \]

Step 6: Determine the maximum value

Since the function \( A(n) = -0.01n^2 + 3n + 7000 \) is a downward opening parabola (the coefficient of \( n^2 \) is negative), the critical point at \( n = 150 \) is a maximum.

Step 7: Calculate the total number of vines

The total number of vines to plant per acre is the original 700 vines plus the additional 150 vines: \[ 700 + 150 = 850 \] vines.

Key Concepts

Quadratic Functions

In algebra, a quadratic function is a polynomial of degree two. The general form of a quadratic function is \( ax^2 + bx + c \), where \( a, b, \) and \( c \) are constants, and \( x \) is the variable. Quadratic functions are recognized by their characteristic parabolic shape when graphed, which can either open upwards or downwards depending on the sign of the coefficient \( a \).

For instance, in the vineyard problem, the function \( A(n) = -0.01n^2 + 3n + 7000 \) is a quadratic function. Here, the coefficient \(-0.01\) denotes the parabola opening downwards, indicating a maximum value at its vertex. The role of quadratic functions in optimization, like maximizing grape production, is crucial as they help determine the optimal point, or maximum yield, in real-world scenarios.

Derivative for Maximum

Derivatives are a fundamental tool in calculus used to determine the rate of change of a function. They are particularly useful in finding maxima and minima, which are crucial in optimization problems.

To find where a quadratic function reaches its maximum or minimum, we use its derivative. Consider the function \( A(n) = -0.01n^2 + 3n + 7000 \). The derivative \( A'(n) = -0.02n + 3 \) gives the slope of the tangent line to the curve at any point \( n \).

Setting this derivative equal to zero helps locate critical points. Here, solving \(-0.02n + 3 = 0\) gives \( n = 150 \), indicating the additional vines needed for maximum grape production. Finding where the derivative is zero helps identify potential maximum or minimum points in the function.

Polynomial Expansion

Polynomial expansion involves expressing a function in an extended form to simplify complex expressions. This is often done to make differentiation and analysis straightforward.

In our exercise, the function \( A(n) = (700+n)(10-0.01n) \) is expanded to \( A(n) = -0.01n^2 + 3n + 7000 \). This expansion uses the distributive property, where each term in the first bracket multiplies every term in the second. The expanded form directly reveals a quadratic structure, which simplifies computation and analysis.

This method is beneficial as it transforms a seemingly complicated product of binomials into a cleaner polynomial form, making further steps like differentiation much easier.

Critical Points in Calculus

Critical points are values of the variable where the derivative of a function is zero or undefined. They are essential in analyzing the function's behavior, particularly in finding local maxima, minima, or points of inflection.

Once the derivative \( A'(n) = -0.02n + 3 \) is obtained, critical points occur where \( A'(n) = 0 \). Solving this equation gives \( n = 150 \), a critical point indicating the maximum yield of grape production. Since it's a quadratic function opening downwards, this critical point is clearly a maximum.

Determining the nature of these points (maxima or minima) can be done by second derivative tests or observing the function's graph. These points are central in ensuring optimal solutions in real-world applications like agriculture.