Show your work to get full credit. Solve the following problem with any method you want.
2 answers:
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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.
9514 1404 393
Answer:
- 1.375 seconds
- 35.25 feet
Step-by-step explanation:
A quadratic function y=ax^2 +bx +c is symmetric about the vertical line x=-b/(2a). That is the x-coordinate of the vertex (maximum or minimum).
Here, we have ...
h = -16t^2 +44t +5
t = -44/(2(-16)) = 11/8 = 1.375
It takes 1.375 seconds for the ball to reach its maximum height.
The height at that time is ...
h = (-16(11/8) +44)(11/8) +5 = (-22 +44)(11/8) +5 = 121/4 +5 = 141/4 = 35.25
The maximum height is 35.25 feet.
