A fence is to be built to enclose a rectangular area of 1800 square feet. The fence along three sides is to
be made of material that costs per foot The material for the fourth side costs per foot. Find the
dimensions of the rectangle that will allow for the most economical fence to be built.
The short side is and the long side is

Ask by Rowe Fitzgerald. in the United States

Mar 24,2025

Question

A fence is to be built to enclose a rectangular area of 1800 square feet. The fence along three sides is to
be made of material that costs per foot The material for the fourth side costs per foot. Find the
dimensions of the rectangle that will allow for the most economical fence to be built.
The short side is and the long side is

Ask by Rowe Fitzgerald. in the United States

Mar 24,2025

UpStudy AI · Accepted Answer

To find the dimensions of the rectangle that will allow for the most economical fence to be built, we need to minimize the total cost of the fence.

Let's denote the short side of the rectangle as $$ x $$ feet and the long side as $$ y $$ feet.

The perimeter of the rectangle is given by:
$$ P = 2x + 2y $$

The cost of the fence is given by:
$$ C = 3(2x + y) + 9y $$

We want to minimize the cost $$ C $$ with respect to $$ x $$ and $$ y $$.

To find the minimum cost, we can use the method of Lagrange multipliers. However, in this case, we can also use the fact that the area of the rectangle is fixed at 1800 square feet:
$$ xy = 1800 $$

We can express $$ y $$ in terms of $$ x $$ as:
$$ y = \frac{1800}{x} $$

Substitute this expression for $$ y $$ into the cost function:
$$ C = 3(2x + \frac{1800}{x}) + 9\frac{1800}{x} $$

Now, we can find the critical points by taking the derivative of $$ C $$ with respect to $$ x $$ and setting it equal to zero:
$$ \frac{dC}{dx} = 3(2 - \frac{1800}{x^2}) - \frac{16200}{x^2} = 0 $$

Solving this equation will give us the value of $$ x $$ that minimizes the cost. Once we have $$ x $$, we can find the corresponding value of $$ y $$ using the area equation $$ xy = 1800 $$.

Let's calculate the derivative and solve for $$ x $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$3\left(2-\frac{1800}{x^{2}}\right)-\frac{16200}{x^{2}}=0$$
- step1: Set the base not equal to 0:
  $$3\left(2-\frac{1800}{x^{2}}\right)-\frac{16200}{x^{2}}=0,x\neq 0$$
- step2: Multiply the terms:
  $$6-\frac{5400}{x^{2}}-\frac{16200}{x^{2}}=0$$
- step3: Multiply both sides of the equation by LCD:
  $$\left(6-\frac{5400}{x^{2}}-\frac{16200}{x^{2}}\right)x^{2}=0\times x^{2}$$
- step4: Simplify the equation:
  $$6x^{2}-21600=0$$
- step5: Move the constant to the right side:
  $$6x^{2}=21600$$
- step6: Divide both sides:
  $$\frac{6x^{2}}{6}=\frac{21600}{6}$$
- step7: Divide the numbers:
  $$x^{2}=3600$$
- step8: Simplify the expression:
  $$x=\pm \sqrt{3600}$$
- step9: Simplify:
  $$x=\pm 60$$
- step10: Separate into possible cases:
  $$\begin{align}&x=60\&x=-60\end{align}$$
- step11: Check if the solution is in the defined range:
  $$\begin{align}&x=60\&x=-60\end{align},x\neq 0$$
- step12: Find the intersection:
  $$\begin{align}&x=60\&x=-60\end{align}$$
- step13: Rewrite:
  $$x_{1}=-60,x_{2}=60$$
The critical points are $$ x = -60 $$ and $$ x = 60 $$. Since the length of a side cannot be negative, we discard $$ x = -60 $$ and consider $$ x = 60 $$ as the critical point.

Now, we can find the corresponding value of $$ y $$ using the area equation $$ xy = 1800 $$:
$$ y = \frac{1800}{x} = \frac{1800}{60} = 30 $$

Therefore, the dimensions of the rectangle that will allow for the most economical fence to be built are:
- Short side: $$ x = 60 $$ feet
- Long side: $$ y = 30 $$ feet

So, the short side is 60 feet and the long side is 30 feet.
The most economical dimensions are 60 feet for the short side and 30 feet for the long side.

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