Problem 4 - Calculus for Biology and Medicine

Question

A rectangular field is bounded on one side by a river and on the other three sides by a fence. Find the dimensions of the field that will maximize the enclosed area if the fence has a total length of \(320 \mathrm{ft}\).

Step-by-Step Solution

Answer

The dimensions of the field are 160 ft by 80 ft for maximum area.

Step 1: Understand the Problem

We are given a rectangular field next to a river, meaning one side of the rectangle doesn't need fencing. We have a total of 320 ft of fencing for the other three sides. Our goal is to find the dimensions of the rectangle that will maximize the area enclosed by the fence.

Step 2: Define Variables

Let the length of the side parallel to the river be \(x\) and the lengths of the other two sides be \(y\). The fencing is used on the two \(y\) sides and one \(x\) side, so we have the constraint \(x + 2y = 320\).

Step 3: Express Area in Terms of One Variable

The area \(A\) of the rectangle is given by \(A = x \times y\). Using the constraint \(x + 2y = 320\), solve for \(y\) in terms of \(x\): \(y = \frac{320 - x}{2}\). Substitute in the area formula to get \(A = x \cdot \frac{320 - x}{2}\) or \(A = 160x - \frac{x^2}{2}\).

Step 4: Find the Maximum Area Using Calculus

To maximize the area, take the derivative of \( A \) with respect to \( x \) and set it to zero: \( \frac{dA}{dx} = 160 - x = 0 \). Solving for \( x \) gives \( x = 160 \).

Step 5: Determine the Value of Y Using X

Substitute \(x = 160\) back into the equation for \(y\): \(y = \frac{320 - 160}{2} = 80\).

Step 6: Verify the Solution

Check if dimensions \(x=160\) and \(y=80\) satisfy the constraint \(x + 2y = 320\). Calculate: \(160 + 2 \times 80 = 320\), which is correct.

Key Concepts

Rectangular Field Problem

The rectangular field problem is a classic optimization problem that often appears in calculus. Imagine you have a piece of land bordered by a river on one side, which means this side requires no fencing. You have a limited amount of fencing available to enclose the other three sides of the field. The challenge is to use the given amount of fencing in such a way that the area enclosed by the fence is maximized.In this particular problem, you have 320 feet of fencing to work with. The fencing will be used on two sides of length \(y\) and one side of length \(x\). The constraint of the problem is therefore \(x + 2y = 320\). Your task is to find the dimensions that will provide the largest possible area for the field using the given fencing constraint.

Maximizing Area

Maximizing the area in an optimization problem involves using a limited resource (in this case, fencing) in the best possible way. The goal is to find the dimensions \(x\) and \(y\) that result in the largest possible area. In terms of formulas, the area \(A\) of the rectangle can be expressed as \(A = x \cdot y\).To simplify the calculation, we express \(y\) in terms of \(x\) using the constraint equation \(x + 2y = 320\). Solving for \(y\), we get \(y = \frac{320 - x}{2}\). Substituting this expression into the area formula, we obtain:

• \(A = x \cdot \frac{320 - x}{2}\)
• or \(A = 160x - \frac{x^2}{2}\)

At this point, the area is expressed as a function of \(x\), and we can proceed to find the value of \(x\) that maximizes \(A\).

Calculus Derivative

Using calculus derivatives is an effective method to find maximum or minimum values of a function, especially in optimization problems. To maximize the area, we first need to take the derivative of the area function \(A = 160x - \frac{x^2}{2}\) with respect to \(x\). The derivative, \(\frac{dA}{dx}\), is found as follows:

• \(\frac{dA}{dx} = 160 - x\)

To find the maximum area, we need to set this derivative equal to zero and solve for \(x\):

• \(160 - x = 0\)
• Solving gives \(x = 160\)

Substitute \(x = 160\) back into the equation for \(y\) to find \(y = \frac{320 - 160}{2} = 80\). With \(x = 160\) and \(y = 80\), you've found the optimal dimensions that maximize the area within the 320 feet of fencing.