Let's solve the given problem step by step.

Problem Setup:

We aim to enclose a pasture using 300 feet of fencing, with one side of the pasture bordered by a river. The side along the river does not require fencing. The objective is to determine the dimensions of the pasture that maximize the area.

Step-by-Step Solution:

(A) Write the Primary Equation:

The Primary Equation is the equation for the area of the pasture: $A = L \cdot W$ Where:

• $L$ = length (parallel to the river)
• $W$ = width (perpendicular to the river)

(B) Write the Secondary Equation:

The total fencing available is 300 feet, and it is used on the three sides of the pasture: $L + 2W = 300$

This is the Secondary Equation, which relates the length and width using the total fencing.

(C) Write the Primary Equation in Terms of One Variable:

Solve the Secondary Equation for $L$: $L = 300 - 2W$

Substitute $L = 300 - 2W$ into the Primary Equation: $A = (300 - 2W) \cdot W$ $A = 300W - 2W^2$

Thus, the Primary Equation in terms of $W$ is: $A = 300W - 2W^2$

(D) Determine the Dimensions that Will Produce a Pasture of Maximum Area:

To find the maximum area, we take the derivative of $A$ with respect to $W$ and set it equal to zero: $\frac{dA}{dW} = 300 - 4W$ Set $\frac{dA}{dW} = 0$: $300 - 4W = 0$ $W = 75$

Now substitute $W = 75$ into the Secondary Equation to find $L$: $L = 300 - 2(75)$ $L = 150$

Thus, the dimensions that maximize the area are: $L = 150 \, \text{feet}, \, W = 75 \, \text{feet}.$

(E) Verify the Maximum Using the Second Derivative Test:

The second derivative of $A$ is: $\frac{d^2A}{dW^2} = -4$

Since $\frac{d^2A}{dW^2} = -4 < 0$, the function $A = 300W - 2W^2$ has a maximum at $W = 75$.

Final Answer:

• Dimensions for Maximum Area: $L = 150$ feet, $W = 75$ feet
• Maximum Area: $A = L \cdot W = 150 \cdot 75 = 11,250 \, \text{square feet}.$

Do you want more explanation or calculations?

Related Questions:

1. What is the significance of using the derivative to maximize the area?
2. Can we confirm these dimensions by graphing the area equation?
3. How would the problem change if all four sides needed fencing?
4. What happens if the total fencing available is less than 300 feet?
5. How does the shape of the pasture affect the area optimization?

Tip:

When solving optimization problems, always clearly identify your primary and secondary equations, as they guide the entire process systematically.