Problem 11 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

A farmer's barn is 60 feet long on one side. He wishes to build a rectangular pen of area 800 square feet along that side of his barn. What should be the dimension of the pen to minimize the amount of fence used?

Step-by-Step Solution

Answer

The pen should be 40 feet long and 20 feet wide to minimize the fence used.

Step 1: Define Variables

Let the length of the pen parallel to the barn be \( x \) and the width be \( y \). We know that one side of the pen is already covered by the barn.

Step 2: Write the Area Equation

The area of the pen is given by the equation: \( xy = 800 \). We need this to express one variable in terms of the other.

Step 3: Express Width in Terms of Length

Solve the area equation \( xy = 800 \) for \( y \) in terms of \( x \): \( y = \frac{800}{x} \).

Step 4: Write the Fence Equation

The total amount of fence used is for the parts not against the barn: \( x + 2y \).

Step 5: Substitute for Width

Substitute \( y = \frac{800}{x} \) into the fence equation: \( f(x) = x + 2\left(\frac{800}{x}\right) \).

Step 6: Simplify the Fence Equation

Simplify to get \( f(x) = x + \frac{1600}{x} \).

Step 7: Find the Derivative of the Fence Function

Find the derivative \( f'(x) = 1 - \frac{1600}{x^2} \).

Step 8: Critical Points

Set \( f'(x) = 0 \) to find critical points: \( 1 - \frac{1600}{x^2} = 0 \). Solve for \( x \): \( x^2 = 1600 \), giving \( x = 40 \), since dimensions can't be negative.

Step 9: Determine Width Using Critical Point

Use \( x = 40 \) in \( y = \frac{800}{x} \) to find \( y = \frac{800}{40} = 20 \).

Step 10: Verify Solution is Minimum

Check the second derivative \( f''(x) = \frac{3200}{x^3} \). Since \( f''(40) = \frac{3200}{64000} \) is positive, the function is minimized at \( x = 40 \).

Step 11: State Final Dimensions

The minimum amount of fence is used when the pen's dimensions are 40 feet by 20 feet.

Key Concepts

Critical Points

Critical points are values in an optimization problem where a function reaches a minima or maxima. They are crucial in determining where a function's output is the smallest or largest, which can help solve an optimization problem efficiently.

In our problem, the objective was to minimize the total amount of fencing used. We used calculus to help find these critical points. First, we defined a function representing the length of fencing needed. The derivative of this function helped us identify the critical points.

• To find the critical points, we set the derivative equal to zero. This gives the points where the slope of the function is zero, indicating potential minima or maxima.
• After solving the derivative equation, we found that at critical point, the length parallel to the barn is 40 feet.

This point is crucial as it helps minimize the cost or resources needed for a project.

Derivative Applications

Derivatives are powerful tools in calculus that allow us to find rates of change. They are also used to locate points where functions hit their turning points. In optimization problems, derivatives help us understand how changing variables affect a function's outcome.

For our exercise, we took the derivative of the function representing the total fence length:

• This derivative was set to zero to discover critical points where the function changes direction.
• By solving the derivative equation, we identified critical values that guided us in minimizing the fence length.

Using the second derivative, we verified the nature of the critical points. A positive second derivative indicates a minimum, confirming that our critical point indeed minimized the fencing required.

Rectangular Area Optimization

Rectangular area optimization involves adjusting the dimensions of a rectangle to achieve desired properties. In various real-life scenarios, we want to either maximize the area, minimize perimeter, or control other parameters within fixed constraints.

In our problem, the farmer needed to minimize the fence length while maintaining a fixed area of 800 square feet for the pen.

• We expressed the area constraint as an equation: \(xy = 800\).
• By manipulating this equation, we expressed width ( y ) in terms of length ( x ), allowing the optimization of the fence length.

We formulated the fencing equation and simplified it to help find the critical points using calculus. This enabled us to find the pen dimensions that used the least amount of fencing, demonstrating practical application of optimization techniques.