The zeros of the function are , 3, and , and the y-intercept of the function is located at .
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Answer:
The length and width of the plot that will maximize the area of the rectangular plot are 54 ft and 27 ft respectively.
Step-by-step explanation:
Given that,
The length of fencing of the rectangular plot is = 108 ft.
Let the longer side of the rectangular plot be x which is also the side along the river side and the width of the rectangular plot be y.
Since the fence along the river does not need.
So the total perimeter of the rectangle is =2(x+y) -x
=2x+2y-y
=x+2y
So,
x+2y =108
⇒x=108 -2y
Then the area of the rectangle plot is A = xy
A=xy
⇒A= (108-2y)y
⇒ A = 108y-2y²
A = 108y-2y²
Differentiating with respect to x
A'= 108 -4y
Again differentiating with respect to x
A''= -4
For maximum or minimum, A'=0
108 -4y=0
⇒4y=108
⇒y=27.
Since at y= 27, A''<0
So, at y=27 ft , the area of the rectangular plot maximum.
Then x= (108-2.27)
=54 ft.
The length and width of the plot that will maximize the area of the rectangular plot are 54 ft and 27 ft respectively.