Question

A farmer has 64 feet of fence to enclose a rectangular vegetable garden. which dimensions would result in the biggest area for this garden

Answer 1

The dimensions that would result to maximum area will be found as follows:
let the length be x, the width will be 32-x
thus the area will be given by:
P(x)=x(32-x)=32x-x²
At maximum area:
dP'(x)=0
from the expression:
P'(x)=32-2x=0
solving for x
32=2x
x=16 inches
thus the dimensions that will result in maximum are is length=16 inches and width=16 inches

Answer 2

The dimensions of the rectangular fencing that farmer should make to encompass greatest area of the garden are 16 feet by 16 feet.

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 the considered case, we have:

• The total length of the fencing = 64 feet
• Fencing will be done in rectangular shape

We need to find the dimensions for which biggest area of the garden can be covered by this fencing.

Let the dimensions be L length and W width.

Then, the perimeter 2(L + W) = length of the fencing = 64 feet

or

$L + W = 64/2 \\\\L = 32 - W \: \rm feet$

The area of this fencing would be:

$A = L\times W = (32 - W)(W) = f(W)$

We need to maximize the value of f(W) for some specific value of W.

Taking its first and second derivative:

$f'(W) = 32 -2W\\f''(W) = -2$

Finding critical values by putting first derivative = 0, we get:

$32 - 2W = 0\\W = 16 \: \rm ft$
Since the second rate = -2 < 0, thus the critical point is maxima, and its maximum value of the function because its the only point of maxima.

Thus, the area maximizes when W = 16 feet.

We get length of the fencing as:

$L = 32 -16 =16 \: \rm ft$

Thus, the dimensions of the rectangular fencing that farmer should make to encompass greatest area of the garden are 16 feet by 16 feet.

Learn more about maxima and minima of a function here:

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