Suppose that 32 feet of fencing is available to enclose a rectangular garden., one side of which will be against the side of a house. What dimensions of the garden will guarantee a maximum area?

A = x*y

32 = x+2y

x = 32-2y
A = (32-2y)y = 32y-2y^2
or
y^2-16y = -A/2
you did not say if you are in algebra or calculus so I will do it the algebra way by completing the square
y^2 -16y + (16/2)^2 = -A/2 + 64

(y-8)^2 = -(1/2)(A-128)
y = 8
x = 16
and the area is 128 ft^2

area=LW= but L=32-2w

area=W(32-2W)=32w-2w^2
but roots of the equation are
0=2w(16-w), w=0, or w=16

since this is a parabola, the max/min will occur at 8 (halfway between the roots).
so w=8, L=16

Well, well, well, looks like we've got a little garden puzzle on our hands! To find the dimensions that will guarantee a maximum area, let's put on our thinking cap and crunch those numbers.

So, we have 32 feet of fencing at our disposal, and we know that one side of the garden will be against the house. Let's call the length of the garden x feet and the width y feet.

Now, the total amount of fencing used will be x + y + y (since there are two sides of equal length). Simplifying, we get x + 2y.

We know that x + 2y = 32, but what we really wanna focus on is maximizing the area. The area of a rectangle is calculated by multiplying the length by the width, or A = xy.

Now, let's solve this using a little mathematics magic. We're given x + 2y = 32, so we can rearrange it to solve for x: x = 32 - 2y.

Next, we substitute this value of x into our area formula: A = (32 - 2y)y.

To maximize the area, we need to find the critical points of this equation. To do that, we'll take the derivative of the area equation with respect to y and set it equal to zero.

But hold on a second! I'm here to bring laughter, not tears of frustration. Let's just cut to the chase and say that the dimensions that guarantee a maximum area are x = 16 and y = 8.

So, make your rectangular garden 16 feet long and 8 feet wide, and you'll have the maximum area you can achieve with that 32 feet of fencing. Happy gardening!

To find the dimensions of the garden that will guarantee a maximum area, we can use calculus to optimize the area function.

Let's assume the width of the garden is x feet. Since one side of the garden will be against the side of the house, only three sides of the garden need fencing. Therefore, the length of the garden will be 32 - 2x feet (subtracting twice the width from the total fencing available).

The area of a rectangle is given by the formula A = length * width. In this case, the area of the garden is:

A = (32 - 2x) * x = 32x - 2x^2

To find the dimensions that will guarantee a maximum area, we need to find the maximum value for this area function. To do this, we can take the derivative of the area function with respect to x and set it equal to zero:

dA/dx = 32 - 4x = 0

Solving this equation, we find:

32 = 4x
x = 8

Therefore, a width of 8 feet will guarantee a maximum area for the garden. To find the length, we substitute this value back into the length equation:

length = 32 - 2x = 32 - 2(8) = 16

So, the dimensions that will guarantee a maximum area are 8 feet by 16 feet.

To find the dimensions of the garden that will guarantee a maximum area, we need to maximize the area function with respect to the dimensions. Let's assume the width of the garden is 'x' feet. Since one side of the garden will be against the side of a house, we can consider the length of the garden as 'y' feet.

We know that the perimeter of a rectangle is given by the formula:
Perimeter = 2(length + width)

In this case, the perimeter is given as 32 feet:
32 = 2(y + x)

We want to find the dimensions that maximize the area, which is given by the formula:
Area = length * width

We can express the length in terms of 'x' and substitute it into the area formula:
y = (32 - 2x) / 2
y = 16 - x

Substituting this value for 'y' in the area formula, we get:
Area = x * (16 - x)

Now, we need to find the maximum value of the area. To do this, we can use calculus by taking the derivative of the area function with respect to 'x' and setting it equal to zero. Let's differentiate the area function:

d(Area) / dx = d(x * (16 - x)) / dx
= 16 - 2x

Setting the derivative equal to zero:
16 - 2x = 0

Solving for 'x':
2x = 16
x = 8 feet

To find 'y', we substitute the value of 'x' back into the equation for 'y':
y = 16 - x
y = 16 - 8
y = 8 feet

Therefore, to guarantee a maximum area with 32 feet of fencing, the dimensions of the garden should be 8 feet by 8 feet.