Problem 5 - Calculus - AP Edition

Question

A rectangular plot of land is to be fenced in using two kinds of fencing. Two opposite sides will use heavy-duty fencing selling for 3 dollars a foot, while the remaining two sides will use standard fencing selling for 2 dollars a foot. What are the dimensions of the rectangular plot of greatest area that can be fenced in at a cost of 6000 dollars?

Step-by-Step Solution

Answer

The dimensions are 500 feet by 750 feet.

Step 1: Understand the problem

We need to find the dimensions of a rectangular plot with maximum area, subject to a constraint on the total cost of the fencing. The total budget for fencing is $6000. Two sides of the rectangle use heavy-duty fencing at $3 per foot, and the other two sides use standard fencing at $2 per foot.

Step 2: Define the variables

Let the length of the rectangle be denoted by \( l \) and the width be \( w \). Assume the heavy-duty fencing is used for the two lengths and the standard fencing for the two widths.

Step 3: Write the cost equation

Since two lengths use the heavy-duty fencing, this costs \( 3(2l) = 6l \). For the two widths, the cost is \( 2(2w) = 4w \). The total cost equation is \( 6l + 4w = 6000 \).

Step 4: Solve the cost equation for one variable

Solve the cost equation for \( w \):\[ 4w = 6000 - 6l \]\[ w = \frac{6000 - 6l}{4} = 1500 - 1.5l \]

Step 5: Write the area equation

The area \( A \) of the rectangle is given by \( A = l \times w \). Substitute \( w \) from the previous step: \[ A = l(1500 - 1.5l) = 1500l - 1.5l^2 \].

Step 6: Find the maximum area using calculus

To find the maximum area, take the derivative of \( A \) with respect to \( l \) and set it to zero: \[ \frac{dA}{dl} = 1500 - 3l = 0 \]. Solving for \( l \) gives \( l = 500 \).

Step 7: Calculate the other dimension

Substitute \( l = 500 \) back into the equation for \( w \): \[ w = 1500 - 1.5(500) = 750 \].

Step 8: Verify the solution

Check that the total cost equation holds: \( 6(500) + 4(750) = 3000 + 3000 = 6000 \), confirming the cost is within the budget. Also, verify that the area is maximized by reassessing the objective function and constraints.

Key Concepts

Calculus Problems

Calculus problems often involve various real-world scenarios where we seek to optimize a particular function under given constraints. These problems typically require a critical understanding of how mathematical concepts apply to tangible situations. In this specific problem, we are tasked with maximizing the area of a rectangular plot while staying within a predefined budget for fencing materials.

• Two types of fencing are used, with different costs.
• We aim to maximize the area of the plot, which requires an understanding of geometry and algebra.
• Optimization problems are common in calculus because they require finding the best solution within a range of possibilities.

Calculus problems challenge students to apply mathematical theories to deduce practical solutions, emphasizing analytical thinking. Understanding these concepts lays the foundation for handling more complex problems in physics, economics, and engineering.

Constraint Optimization

Constraint optimization involves finding the best solution within a set of boundaries or limitations. In calculus, this often means maximizing or minimizing a function while respecting constraints. In the example problem, we have a budget constraint, which dictates how we allocate resources to maximize the area.

• We define variables to represent the lengths and widths of the plot.
• The cost equation integrates constraints: heavy-duty fencing costs more, imposing a limit on the perimeter.
• Using algebra, we express width in terms of length, incorporating the budget constraint.

We achieve optimization by reshaping our primary equation to reflect the constraint. This methodical approach ensures resourceful and efficient problem-solving, critical for scenarios where resources are finite.

Differentiation Techniques

Differentiation is a cornerstone of calculus that helps find rates of change and function extremities. In the context of this problem, we use differentiation to find the maximum area of the plot.
To solve the problem:

• We express the area as a function in terms of one variable, here the length.
• Then, we take the derivative of this function with respect to the variable.
• Setting the derivative to zero helps locate the function's critical points — potential maxima or minima.

The insight gleaned from setting the derivative to zero is that the slope of the tangent is momentarily flat, suggesting a peak or a trough in the function's graph. By further evaluating these points, particularly with secondary tests if necessary, we can conclusively determine that the plot dimensions lead to a maximum area under the given budgetary constraint.