A rectangular-shaped garden has one side along the side of a house. the other three sides are to be enclosed with 60ft of fencing. what is the largest possible area of such a garden?

x = length

y = width
A = x y
P = 60 = x + 2 y
so
x = 60 - 2 y

A = (60 - 2 y) y = 60 y - 2 y^2

2 y^2 - 60 y = -A
y^2 - 30 y = -A/2
complete square for vertex of parabola
y^2 - 30 y + 225 = -A/2 + 225
(y-15)^2 = -A/2 + 225 = -(1/2)(A-450)
so y = 15 at vertex
A = 450 at vertex
so
15 by 30 for 450 ft^2 area

To find the largest possible area of a rectangular garden, we need to consider that the perimeter of the garden is fixed at 60ft.

Let's assume that the width of the garden is x ft. Since one side of the garden is against a house, the length of the garden will be 60 - 2x ft (subtracting twice the width from the fixed perimeter).

The area of a rectangle is calculated by multiplying its length and width. Therefore, we need to find the width that maximizes the area.

Area (A) = Length (l) * Width (w)
A = (60 - 2x) * x

To find the maximum area, we can take the derivative of the area function with respect to x, set it equal to zero, and solve for x.

dA/dx = 0
2x - 60 = 0

Solving for x gives us x = 30ft.

Since the width cannot be greater than half the total length (30ft), we can conclude that the largest possible area of the garden is when x = 30ft.

Plugging this value back into the area formula:
A = (60 - 2(30)) * 30
A = 30 * 30
A = 900 square feet

Therefore, the largest possible area of such a garden is 900 square feet.