Question

Minimum cost A rectangular area adjacent to a river is to be fenced in; no fence is needed on the river side. The enclosed area is to be 1000 square feet. Fencing for the side parallel to the river is \(\$ 5\) per linear foot, and fencing for the other two sides is \(\$ 8\) per linear foot; the four corner posts are \(\$ 25\) apiece. Let \(x\) be the length of one of the sides perpendicular to the river. (a) Write a function \(C(x)\) that describes the cost of the project. (b) What is the domain of \(C ?\) (c) Use a graphing utility to graph \(C=C(x)\). (d) Find the dimensions of the cheapest enclosure.

Short answer

Dimensions for the cheapest enclosure are approximately 17.68 feet by 56.59 feet.

Step-by-step solution

- Define Variables and Given Information

Let's define the variables. Let x be the length of one side perpendicular to the river, and let y be the length parallel to the river. The enclosed area is 1000 square feet. So, we have the equation: \[ x \times y = 1000 \].

- Express y in terms of x

From the area equation, express y in terms of x: \[ y = \frac{1000}{x} \].

- Write the cost function

The total cost includes the cost of the fencing for three sides and the four corner posts. The cost function is: \[ C(x) = 5y + 16x + 100 \]. Substituting y: \[ C(x) = 5 \left( \frac{1000}{x} \right) + 16x + 100 = \frac{5000}{x} + 16x + 100 \].

- Determine the Domain of C(x)

Since x must be a positive length, but also y must be positive, the domain of C(x) is \( 0 < x < \infty \).

- Graph the Cost Function

Use graphing software or a graphing utility to graph the cost function \( C(x) = \frac{5000}{x} + 16x + 100 \). Observe the behavior of the function to find the minimum point.

- Find the Minimum Cost and Corresponding Dimension

To find the minimum cost, take the derivative of C(x) and set it to zero: \[ C'(x) = -\frac{5000}{x^2} + 16 \]. Solve for x when \( C'(x) = 0 \): \[ -\frac{5000}{x^2} + 16 = 0 \]. This simplifies to: \[ 16x^2 = 5000 \]. Solving for x: \[ x^2 = \frac{5000}{16} = 312.5 \]. Therefore, \( x = \sqrt{312.5} \approx 17.68 \). Calculate y as \[ y = \frac{1000}{x} \approx 56.59 \].

- Verify the Minimum

Use the second derivative test to verify the minimum. The second derivative \( C''(x) = \frac{10000}{x^3} \) is positive for all \( x > 0 \), confirming a minimum at \( x \approx 17.68 \).

Key concepts

Cost Function

In this exercise, the cost function represents the total cost of fencing for a rectangular area beside a river. To create the cost function, we consider the cost of materials and the dimensions provided. The side parallel to the river costs \$5 per linear foot, while the sides perpendicular to the river cost \$8 per linear foot each. Finally, there are four corner posts costing \$25 each.
Using these rates, we can express the cost function as follows: If \(x\) is the length of the side perpendicular to the river, and \(y\) is the length parallel to the river, the area \(xy = 1000\) square feet.
Simplifying, we express \(y\) in terms of \(x\): \(y = \frac{1000}{x}\).
Therefore, the cost function (\(C(x)\)) is:
\[ C(x) = 5y + 16x + 100 \ \Rightarrow C(x) = 5 \left( \frac{1000}{x} \right) + 16x + 100 \ \Rightarrow C(x) = \frac{5000}{x} + 16x + 100 \].

Derivatives

Derivatives help us find the minimum cost by locating where the cost function’s rate of change is zero. First, take the derivative of the cost function \(C(x) = \frac{5000}{x} + 16x + 100\):
\[ C'(x) = -\frac{5000}{x^2} + 16 \].
The derivative \(C'(x)\) tells us the slope of the cost function at any point \(x\). To find where this function has a minimum, we set the derivative equal to zero:
\[ -\frac{5000}{x^2} + 16 = 0 \ \Rightarrow 16x^2 = 5000 \ \Rightarrow x^2 = 312.5 \ \Rightarrow x = \sqrt{312.5} \approx 17.68 \].
This calculation highlights that the cost function has a critical point at \(x \approx 17.68\).

Domain and Range

The domain of a function includes all possible values of the input variable that make the function valid. For our cost function \(C(x)\), \(x\) represents the length of one side perpendicular to the river. Both \(x\) and \(y\) (the side parallel to the river) must be positive because negative lengths don't make sense
The function’s domain is therefore all positive values of \(x\): \( 0 < x < \infty \).
For the range of \(C(x)\), observe the values of the cost function. Given the cost expression \(C(x) = \frac{5000}{x} + 16x + 100\), the total cost will be positive because each part \(\frac{5000}{x}, 16x, \text{and} 100\) is positive for positive values of \(x\).

Optimization

Optimization involves finding the minimum or maximum value of a function. Here, we aim to minimize the cost of fencing. We achieved this by finding the critical point using the first derivative.
To confirm the critical point at \(x \approx 17.68\) is a minimum, we use the second derivative test. The second derivative of the cost function \(C(x) = \frac{5000}{x} + 16x + 100\) is:
\[ C''(x) = \frac{10000}{x^3} \].
Since \(C''(x)\) is positive for all \(x > 0\), the cost function is concave up, indicating a minimum at \(x \approx 17.68\).
Finally, we calculate the corresponding \(y\) value:
\[ y = \frac{1000}{x} \approx \frac{1000}{17.68} \approx 56.59 \].
Thus, the dimensions that minimize cost are approximately 17.68 feet perpendicular to the river and 56.59 feet parallel to the river.