Question Statement

If the total fence to be used is 8000 m, find the dimensions of the enclosed land in the figure below that has the greatest area.

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Background and Explanation

This problem involves maximizing the area of a rectangular enclosure subject to a fixed constraint on the total fencing available, using techniques from differential calculus. The key approach is to express the area as a function of a single variable using the perimeter constraint, then identify critical points where the first derivative equals zero and verify they represent a maximum using the second derivative test.

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Solution

Setting Up the Constraint

Let $x$ be the length of the rectangle and $y$ be the width. Based on the figure, the fencing arrangement requires two lengths of side $x$ and four lengths of side $y$ (this configuration typically represents a rectangular field with internal dividers parallel to the width).

The total fencing constraint is:

$\begin{array}{rl}\text{Perimeter}& =x+x+y+y+y+y\\ \text{Perimeter}& =2x+4y\end{array}$

Given that the total fence available is 8000 m:

$2x+4y=8000$

Simplifying by dividing by 2:

$x+2y=4000$

Solving for $x$ in terms of $y$:

$x=4000-2y$

Expressing the Area Function

The area of the rectangular enclosure is given by:

$A=x\times y$

Substituting equation (1) to express the area solely in terms of $y$:

$\begin{array}{rl}A& =(4000-2y)\times y\\ A& =4000y-2y^2\end{array}$

Finding Critical Points

To find the value of $y$ that maximizes the area, we differentiate $A$ with respect to $y$:

$\frac{dA}{dy}=4000-4y$

We also compute the second derivative for the concavity test:

$\frac{d^{2}A}{d{y}^2}=-4$

Setting the first derivative equal to zero to find critical points:

$\begin{array}{rl}4000-4y& =0\\ 4000& =4y\\ y& =1000\end{array}$

Verifying the Maximum

Since the second derivative is:

$\frac{d^{2}A}{d{y}^2}=-4<0$

The negative value confirms that the function is concave down at $y=1000$, indicating that this critical point yields a local maximum for the area.

Calculating the Final Dimensions

Substitute $y=1000$ back into equation (1) to find $x$:

$\begin{array}{rl}x& =4000-2(1000)\\ & =4000-2000\\ x& =2000\end{array}$

Therefore, the dimensions that yield the greatest enclosed area are:

• Length: $x=2000$ m
• Width: $y=1000$ m

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Key Formulas or Methods Used

• Perimeter constraint: $2x+4y=8000$ (simplified to $x=4000-2y$)
• Area function: $A=xy$ or $A=4000y-2y^2$
• First derivative (optimization): $\frac{dA}{dy}=4000-4y$
• Critical point condition: $\frac{dA}{dy}=0$
• Second derivative test for maximum: $\frac{d^{2}A}{d{y}^2}<0$ confirms concave down (maximum)

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Summary of Steps

1. Define variables $x$ (length) and $y$ (width) for the rectangular enclosure.
2. Establish the fencing constraint based on the figure: $2x+4y=8000$.
3. Solve the constraint for $x$ to get $x=4000-2y$.
4. Substitute into the area formula to obtain $A(y)=4000y-2y^2$.
5. Differentiate with respect to $y$ to get $\frac{dA}{dy}=4000-4y$.
6. Set the first derivative to zero and solve: $y=1000$ m.
7. Verify this is a maximum using the second derivative test ($\frac{d^{2}A}{d{y}^2}=-4<0$).
8. Substitute $y=1000$ back to find $x=2000$ m and state the final dimensions.