A Man Wants To Build A Rectangular Enclosure
A Man Wants To Build A Rectangular Enclosure For
A man wants to build a rectangular enclosure for his herd. He has a budget of $900 to spend on fencing and aims to maximize the size of the enclosure. The enclosure will be built along a river, so one side does not require fencing. The fencing cost per foot along the side parallel to the river is $5, while the two perpendicular sides cost $3 per foot. The task is to determine the optimal dimensions of the enclosure that maximize its area within the given budget constraints.
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The problem involves optimizing the size of a rectangular enclosure with specific cost constraints and partial fencing along a river. To find the dimensions that yield the maximum area within a $900 budget, it is necessary to formulate the problem mathematically and then apply optimization techniques under given constraints.
Assuming the length of the side along the river is represented by \(L\), and the widths of the perpendicular sides are represented by \(W\), the total fencing used comprises the non-river sides: the two widths and the side opposite the river. Since fencing is only needed for three sides, the total fence length \(F\) is:
\[F = 2W + L\]
The cost of fencing, considering the differing prices per foot, is:
\[
\text{Cost} = 3 \times (2W) + 5 \times L = 6W + 5L
\]
Given the total budget of \(\$900\), the cost constraint becomes:
\[
6W + 5L \leq 900
\]
The area \(A\) of the enclosure, which we aim to maximize, is:
\[A = L \times W\]
Since the material cost constraint involves both \(L\) and \(W\), this is a constrained optimization problem. To proceed, express \(L\) in terms of \(W\) from the cost constraint:
\[
L \leq \frac{900 - 6W}{5}
\]
To maximize the area, consider the equality case (since a larger \(L\) leads to a larger area) and substitute:
\[
A(W) = W \times \frac{900 - 6W}{5}
\]
\begin{equation}
A(W) = \frac{W(900 - 6W)}{5} = 180W - \frac{6}{5} W^2
\end{equation}
Differentiating \(A(W)\) with respect to \(W\) to find critical points:
\[
\frac{dA}{dW} = 180 - \frac{12}{5} W
\]
Setting derivative to zero:
\[
180 - \frac{12}{5} W = 0 \Rightarrow \frac{12}{5} W = 180 \Rightarrow W = \frac{180 \times 5}{12} = 75
\]
Then, find \(L\):
\[
L = \frac{900 - 6 \times 75}{5} = \frac{900 - 450}{5} = \frac{450}{5} = 90
\]
Therefore, the optimal dimensions are:
- Width: 75 feet
- Length along the river: 90 feet
The maximum area achievable is:
\[
A_{max} = 75 \times 90 = 6750 \text{ square feet}
\]
The total fencing cost at these dimensions is:
\[
6 \times 75 + 5 \times 90 = 450 + 450 = 900
\]
which exactly matches the budget.
In conclusion, to maximize the area of the enclosure within a \$900 fence budget, the man should build a 75-foot wide side perpendicular to the river and a 90-foot side along the river, resulting in a maximum area of 6,750 square feet.
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