SOLUTION: This is one is tough!
A rectangular enclosure is made with 100 ft. of fencing on three sides. The fourth side is the wall of a barn. Find the greatest possible area of such an
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A rectangular enclosure is made with 100 ft. of fencing on three sides. The fourth side is the wall of a barn. Find the greatest possible area of such an
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Question 174569: This is one is tough!
A rectangular enclosure is made with 100 ft. of fencing on three sides. The fourth side is the wall of a barn. Find the greatest possible area of such an enclosure. My choices are as follows: A) 400ft^2 B) 625 ft^2 C) 1111.1 ft ^2 D) 2500ft^2. Found 2 solutions by nerdybill, Edwin McCravy:Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! If you only have 100 feet of fencing, this means that the perimeter has a maximum of 100 feet.
.
Let x = width
and y = length
.
Perimeter = 2x+y
100 = 2x+y
solving for y:
y = 100 - 2x
.
Area = xy
substituting in our value of y:
Area = x(100 - 2x)
Area = -2x^2 + 100x
.
This is essentially a parabola that opens downward (from the coefficient -2). So, all we need to do is to find the vertex to find the maximum.
.
The x coordinate = -b/2a
The x coordinate = -(100)/2(-2)
The x coordinate = (-100)/(-4)
The x coordinate = (100)/(4)
The x coordinate = 25 feet (width)
.
Length is:
y = 100 - 2x
y = 100 - 2(25)
y = 100 - 50
y = 50 feet (length)
.
Area is then:
25(50) = 1250 square feet
You can put this solution on YOUR website! Edwin's solution:
A rectangular enclosure is made with 100 ft. of fencing on three sides. The fourth side is the wall of a barn. Find the greatest possible area of such an enclosure. My choices are as follows: A) 400ft^2 B) 625 ft^2 C) 1111.1 ft ^2 D) 2500ft^2.
Let y = the area.
The vertex of the parabola
is , so the vertex here is ,
The graph is
So the maximum area is . That is not one of the
choices, so the problem is botched somehow.
Edwin
