Isaac wants to enclose his backyard with a fence that is bounded on one side by the back of the house. if he has 164 feet of fencing material find the maximum area that can be enclosed by the fence.

let the side parallel to the house be y ft

let each of the other two sides by x ft
so
2x + y = 164
y = 164 - 2x

area = xy
= x(164 - 2x
= -2x^2 + 164x

This is a downwards opening parabola, so we need its vertex.
Either
1. complete the square
2. use Calculus
3. use " the x of the vertex is -b/(2a) for any y = ax^2+bx+c

Iwill use the third method:
x of the vertex = -164/-4 = 41
then y = 164-82 = 82

the maximus area = xy = (41)(82) = 3362 square ft
when the field is 41 by 82 ft

To find the maximum area that can be enclosed by the fence, we need to determine the dimensions of the fence that result in the largest area. Since one side of the fence is bounded by the back of the house, the other three sides form a rectangle.

Let's assume the length of the rectangle is x feet. The width, then, will be (164 - x)/2 feet since there are two equal sides left for the width.

The area enclosed by the rectangle is given by the formula A = length * width.
A = x * (164 - x)/2

To find the maximum area, we can take the derivative of the area equation with respect to x and set it equal to zero to find the critical points:

dA/dx = (164 - 2x)/2

Setting this derivative equal to zero, we get:

(164 - 2x)/2 = 0
164 - 2x = 0
2x = 164
x = 82

This means the critical point occurs at x = 82 feet.

Now, to determine if this point is a maximum, minimum, or neither, we can take the second derivative of the area equation:

d^2A/dx^2 = -2/2
d^2A/dx^2 = -1

Since the second derivative is negative, this means that the critical point at x = 82 is a maximum.

Now, we can substitute the value of x = 82 back into the area equation to find the maximum area:

A = 82 * (164 - 82)/2
A = 82 * 82/2
A = 3364 square feet

Therefore, the maximum area that can be enclosed by the fence is 3364 square feet.