A farmer has 20 yards of fencing to build a pen for her chickens. She decides to use a side of her barn as one side of the fenced-in area. What is the maximum area she can achieve?

The farmer can achieve a maximum area of 50 square yards with 20 yards of fencing.

Step-by-step explanation:

Given that farmer shall construct a rectangular fenced-in area and a side of the barn is one side of such area, the needed length of fencing is represent by the following perimeter equation (), measured in square yards:

Where:

- Length of the rectangle, measured in yards.

- Width of the rectangule (side of the barn), measured in yards.

In addition, the equation of the fenced-in area () is:

If , equation of area is now simplified as follows:

The value of associated with the maximum area is obtained with the help of First and Second Derivative Tests. Firstly, first and second derivatives of the area function are determined:

Let equalize first equation to zero, second derivative indicates that critical value follows to an absolute maximum. Hence:

The width of the rectangle is: ( and )

And finally, the maximum area she can achieve is:

The farmer can achieve a maximum area of 50 square yards with 20 yards of fencing.

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

-1/3(x+1)^5

i’m pretty sure this is correct