A veterinarian uses 300 feet of chain-link fencing to enclose a rectangular region and also to subdivide the region into two smaller rectangular regions by placing a fence parallel to one of the sides,


(a) Write the width w as a function of the length l.
w = 1

(b) Write the total area A as a function of l.
A = 2

(c) Find the dimensions that produce the greatest enclosed area.
3 ft (smaller value)
4 ft (larger value)

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(a) To write the width w as a function of the length l, we need to consider that the total length of the fencing is 300 feet. The perimeter of a rectangle is calculated by adding the lengths of all four sides. In this case, since we are dividing the region into two smaller rectangles with a fence parallel to one of the sides, we need to subtract the length of the dividing fence.

The perimeter P of the rectangle can be expressed as:

P = 2w + 2l - x

where w is the width, l is the length, and x is the length of the dividing fence.

Since the total length of the fencing is 300 feet, we can write:

300 = 2w + 2l - x

We can rearrange this equation to solve for w:

2w = 300 - 2l + x

w = (300 - 2l + x)/2

(b) To write the total area A as a function of l, we can use the formula for the area of a rectangle:

A = w * l

Substituting the expression for w from part (a), we have:

A = [(300 - 2l + x)/2] * l

Simplifying this expression may provide a clearer representation based on the given information.

(c) To find the dimensions that produce the greatest enclosed area, we can use calculus by finding the critical points of the area function. However, since the values of w and x are not specified, we cannot provide an exact solution for the dimensions that maximize the area.

To answer parts (a) and (b), we need to understand the given information.

(a) We are told that a veterinarian uses 300 feet of chain-link fencing to enclose a rectangular region and divide it into two smaller rectangular regions. This means we have to find a way to express the width of the rectangular region in terms of the length.

Assuming we have a rectangle with width w and length l, we can identify three sides that use the fencing: two sides with length w and one side with length l. Since there are two smaller rectangular regions, we can represent the lengths of their sides as w and (l - 2w), respectively (because the dividing fence reduces the length by twice the width).

To get the total amount of fencing, we sum up the three sides: w + w + (l - 2w) = 2w + (l - 2w) = l.

Thus, we have the equation 2w + (l - 2w) = l, which simplifies to l - 2w = l. By rearranging terms, we get -2w = -w + l, and finally, w = l.

So, we can express the width w as a function of the length l as w = l.

(b) To write the total area A as a function of the length l, we need to recall that the area of a rectangle is given by multiplying its length and width: A = l * w.

Substituting the expression for w from part (a), we have A = l * (l) = l^2.

Hence, the total area A can be expressed as a function of l as A = l^2.

To answer part (c), we need to find the dimensions that produce the greatest enclosed area. In other words, we need to find the maximum value of the area function we obtained in part (b).

By taking the derivative of A = l^2 with respect to l, we can find the critical points. Setting the derivative equal to zero, we have 2l = 0.

Solving for l, we find that l = 0. Since we cannot have a negative or zero length, we discard this solution.

Therefore, there are no critical points, and we need to analyze the endpoints of the interval.

Given that the veterinarian has 300 feet of chain-link fencing, we can use the perimeter equation to find the range of possible lengths: 2w + 2(l - 2w) = 300.

Simplifying, we get 2w + 2l - 4w = 300, which results in 2l - 2w = 300. By rearranging terms, we obtain 2l = 2w + 300, and finally, l = w + 150.

Since we know that w = l from part (a), we can substitute that into the equation to get l = l + 150.

Simplifying further, we find 0 = 150, which is not possible. Hence, no lengths result in a maximum enclosed area.

Therefore, there are no dimensions that produce the greatest enclosed area in this scenario.