# Ian wants to fence three sides of a rectangular exercise yard for his dog. The fourth side of the exercise yard will be a side of the house. He has 100 feet of fencing available. Find the dimensions that will enclose the maximum area.

## To find the dimensions that will enclose the maximum area, let's denote the width of the exercise yard as "w" and the length as "l".

Since there are three sides to be fenced, we have the following equation for the perimeter of the exercise yard:
2w + l = 100 (Equation 1)

To maximize the area, we need to express the area in terms of a single variable, either "w" or "l". Since the fourth side of the exercise yard is the house, the length "l" will be constant and equal to the length of the house.

Now, we can express the area A of the exercise yard as:
A = w * l

Substituting the value of "l" from Equation 1 into the area equation, we get:
A = w * (100 - 2w)

Expanding this equation, we have:
A = 100w - 2w^2

To find the dimensions that will enclose the maximum area, we need to find the value of "w" that maximizes the area.

Next, we can find the derivative of the area function with respect to "w":
dA/dw = 100 - 4w

Setting the derivative equal to zero to find the critical points:
100 - 4w = 0
4w = 100
w = 25

To determine if this point gives a maximum or minimum, we can take the second derivative:
d^2A/dw^2 = -4

Since the second derivative is negative, this implies that the point "w = 25" is a local maximum.

Therefore, the width of the exercise yard that will enclose the maximum area is 25 feet.

Substituting this value back into Equation 1, we can solve for the length:
2w + l = 100
2(25) + l = 100
50 + l = 100
l = 50

Therefore, the dimensions that will enclose the maximum area are a width of 25 feet and a length of 50 feet.

## To find the dimensions that will enclose the maximum area, let's assume the length of the rectangular exercise yard is L and the width is W.

Given that Ian has 100 feet of fencing available, we know that the total length of the fence used will be:

2L + W = 100

We also know that one side of the fence is the side of the house, so we have:

L + W = 50

To solve this system of equations, we can use substitution or elimination method.

Using substitution method, we can solve the second equation for L:

L = 50 - W

Substituting this value of L into the first equation, we get:

2(50 - W) + W = 100

Simplifying the equation, we have:

100 - 2W + W = 100

-2W + W = 100 - 100

-W = 0

W = 0

Since we can't have a width of 0, this solution is not valid.

Therefore, let's try the elimination method to solve the system of equations:

First, multiply the second equation by 2:

2L + 2W = 100

Next, subtract the first equation from this new equation:

(2L + 2W) - (2L + W) = 100 - 50

2L + 2W - 2L - W = 50

W = 50

Now, substitute this value of W into the second equation to find L:

L + 50 = 50

L = 0

Again, we can't have a length of 0, so this solution is not valid.

After exploring both methods, we can conclude that there are no valid dimensions for the rectangular exercise yard that will enclose the maximum area using 100 feet of fencing.

## 50 by 25

As with all such problems, the maximum area is when the fence is divided equally among lengths and widths.

x+2y = 100
area = xy = x(100-2x)
now just find the vertex of that parabola.