. What is the area of the new enclosure A(w) in terms of width, w? b. What is the maximum area of the new enclosure? What are the dimensions?

Question

. A camp wants to create a larger space for their albino rabbit, Clover. They want to reuse the materials from Clover’s current enclosure in the construction of a new rectangular enclosure. The perimeter of Clover’s current space is 6 feet. The perimeter of his new enclosure will be 3 times larger than his former enclosure.

this is the description

Answer 1

If you want to solve the problem in real numbers, Here it is how:

The new perimeter is 3x6 =18 feet. Let w be the new width, then the new length is (9-w), so the new area is

A(w) = (9-w)w = -w^2 + 9w.

A(w) = -w^2 + 9w is a quadratic function, it reaches maximum at w = -b/(2a), where b is the coefficient of the linear term, a is the coefficient of the the quadratic term. So, w = -b/(2a) = -9/[2x (-1)]} =4.5ft .

The maximum area is 4.5(9-4.5) = 20.25 square ft

the length = the width = 4.5 ft.

Answer 2

As Mark said it does look like this problem is missing some information, however if I make some assumptions I can show you the process to solve it.

A)     If the perimeter of the original enclosure was 6 feet, we know that 2L + 2W = 6. So, if the new enclosure is going to have a perimeter 3 times larger we know that 2L + 2W = 3(6) or

2L + 2 W = 18.

To solve for the area of the new enclosure A(W) in terms of W, you need to find an equation for L in terms of W.

2L + 2W = 18

L + W = 9 (divide equation by 2)

L = 9 – W (subtract W from each side)

Now you can plug into the formula for area, A = LW

A(W) = W * (9 – W)

A(W) = 9W – W^2 (distribute the W)

B)     This is the part where I need to make an assumption. In this case I am going to assume that the length and width can only be whole numbers. If that is the case the possible combinations of length and width are:

1 and 8

2 and 7

3 and 6

4 and 5

(L + W = 9)

Now figure out what the area of each of these combinations would be:

1 * 8 = 8

2 * 7 = 14

3 * 6 = 18

4 * 5 = 20

So the maximum possible area is 20 feet with dimensions 4*5.