Question

Suppose that you want to enclose a rectangular area at the back of your house using the house as one side of the enclosure and fencing on the other three sides. If you have a total of 320 feet of fencing to use for this task, what are the dimensions that will maximize the area and what is the maximum area that can be enclosed?

Answer 1

Answer:

The dimensions of the rectangle for maximum area are 160 feet by 80 feet.

The maximum area enclosed is 12800 square feet.

Step-by-step explanation:

Let the length of the rectangular fencing be 'l' and width be 'w'.

Given:

Length of fencing available = 320 feet

The fencing is done such that one side is covered by the house and the other 3 sides by the fencing available.

Therefore, the perimeter of the rectangle will be equal to the length of fencing available and the width of the house.

This can be written as:

Perimeter = $320 + l$  ( 'l' is the width of house covering one side)

Now, we know that, perimeter of a rectangle is given as:

Perimeter = $2l+2w$

Therefore, $320+l=2l+2w$

$l+2w=320\\l=320-2w---- (1)$

Now, area of a rectangle is given as:

$A=l\times w$

Plug in the value of 'l' from equation (1). This gives,

$A=(320-2w)\times w\\A=320w-2w^2$

Now, for maximum area, the derivative of 'A' must be 0. So, differentiating the above equation with respect to 'w'. This gives,

$\frac{dA}{dw}=320-2\times 2\times w\\\frac{dA}{dw}=320-4w$

Now, for maximum area, $\frac{dA}{dw}=0$. So,

$320-4w=0\\4w=320\\w=\frac{320}{4}=80\ ft$

Also, $l=320-2w=320-2(80)=320-160=160\ ft$

Therefore, the length of the rectangle is 160 feet and width is 80 feet for maximum area.

The value of maximum area is given as:

$A=lw\\\\A=160\times 80\\\\A=12800\ ft^2$

Hence, the maximum area enclosed is 12800 square feet for the dimensions 160 feet by 80 feet.

Answer 2

The dimensions that will maximize the area are: 80 ft and 160 ft . The maximum area that can be enclosed is 12800 sq. ft

How to obtain the maximum value of a function?

To find the maximum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.

Putting those values of x in the second rate of function, if results in negative output, then at that point, there is maxima. If the output is positive then its minima and if its 0, then we will have to find the third derivative (if it exists) and so on.

For this case, we've got:

One side of the rectangle = length of one wall of house = W ft  (let it be width of the rectangle).

Then, let the length of the rectangle be denoted by: L ft.

Then, as the fencing needs to be done on 3 sides only, including 2 lengths and 1 width (as 1 width is closed by wall of house).

Thus, $320 = 2L + W$(as fencing is of 320 ft, and only 3 sides to be covered)

Now, the area of the rectangle is $A = L \times W$

Using the first equation, we get expression of W as:

$W = 320 - 2L \: \rm ft$

Putting this in formula for area as:

$A = L \times W = L \times (320 - 2L) = 320L - 2L^2$

Let we denote $f(L) =320L - 2L^2$

Then, the maximum area is the maximum value of this function.

Finding the first and second rate of this function, we get:

$f'(L) = 320 - 4L\\f''(L) = -4$

Thus, all critical points will be maxima as second rate is negative.

For critical points, putting first rate of the function equal to 0, we get:

$320 - 4L =0 \\L = 80$(in feet)

Thus, at this length, the area of that rectangular region comes maximum.

Width at this length evaluates to: $W = 320 - 2L = 320 -2(80) = 160 \: \rm ft$

Thus, the area of the considered rectangular region comes out as: $A = L \times W = 80 \times 160 = 12800 \: \rm ft^2$

Thus, the dimensions that will maximize the area are: 80 ft and 160 ft . The maximum area that can be enclosed is 12800 sq. ft

Learn more about maxima and minima of a function here:

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