Question

A rancher with 850 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle (see the figure). Find the largest possible total area (in ft2) of the four pens.

Answer 1

The largest possible total area of the four pens is approximately
`\( 30, 104.2 \, \text{ft}^2 \).`

How did we get the value?

To solve this problem, let's denote the dimensions of the rectangular area as follows:

-
`\( x \)`: The width of the rectangular area.

-
`\( y \)`: The length of the rectangular area.

The farmer has 850 ft of fencing, and the fencing consists of four sides. Since the rectangle is divided into four pens with fencing parallel to one side, the total length of the fencing is given by:

`\[ 2x + 3y = 850 \]`

Now, we want to express the area
`\( A \)` in terms of
`\( x \)` and
`\( y \)` and then find the maximum value of
`\( A \)`.

The total area of the four pens is the product of the width and length of the rectangular area:

`\[ A = xy \]`

We can express
`\( y \)` in terms of
`\( x \)` using the equation
`\( 2x + 3y = 850 \)`:

`\[ 3y = 850 - 2x \]`

`\[ y = (850 - 2x)/(3) \]`

Now, substitute this expression for
`\( y \)` into the equation for
`\( A \)`:

`\[ A(x) = x \left( (850 - 2x)/(3) \right) \]`

Now, let's find the critical points by taking the derivative of
`\( A(x) \)` with respect to
`\( x \)` and setting it equal to zero:

`\[ (dA)/(dx) = (850 - 4x)/(3) \]`

Setting this equal to zero:

`\[ 850 - 4x = 0 \]`

Solving for
`\( x \)`:

`\[ x = (850)/(4) = 212.5 \]`

Now, we need to check the second derivative to confirm that this critical point corresponds to a maximum. Taking the second derivative:

`\[ (d^2A)/(dx^2) = -(4)/(3) \]`

Since the second derivative is negative, the critical point
`\( x = 212.5 \)`corresponds to a maximum.

Now, we can find the corresponding
`\( y \)` by substituting
`\( x = 212.5 \)` into the original equation:

`\[ 2x + 3y = 850 \]`

`\[ 2(212.5) + 3y = 850 \]`

`\[ 425 + 3y = 850 \]`

`\[ 3y = 425 \]`

`\[ y = (425)/(3) \]`

Now, we have the dimensions of the rectangle that maximizes the total area of the four pens:

`\[ x = 212.5 \, \text{ft} \]`

`\[ y = (425)/(3) \, \text{ft} \]`

To find the maximum area, substitute these values back into the original equation for
`\( A \)`:

`\[ A(212.5) = 212.5 \left( (850 - 2(212.5))/(3) \right) \]`

`\[ A(212.5) = 212.5 \left( (425)/(3) \right) \]`

`\[ A(212.5) = (212.5 * 425)/(3) \]`

`\[ A(212.5) \approx 30,104.2 \, \text{ft}^2 \]`

So, the largest possible total area of the four pens is approximately
`\( 30, 104.2 \, \text{ft}^2 \).`

Complete question:

Consider the following problem: A farmer with 850 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens? (a) Draw several diagrams illustrating the situation, some with shallow, wide pens and some with deep, narrow pens. Find the total areas of these configurations. Does it appear that there is a maximum area? If so, estimate it.