a rectangular dog pen is to be constructed so that one side is against an existing stone wall and the other three sides are to be fenced. of 500 feet of fence is to be used, determine the dimensions and area of the pen with maximum area.
I have no idea how to solve this problem.
3 answers
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The max area is always a square.
You only need the fence for 3 sides, so divide 500/3
166.67 x 166.67 = max. area.
You only need the fence for 3 sides, so divide 500/3
166.67 x 166.67 = max. area.
The maximum area occurs when the fencing is divided equally among lengths and widths. For all 4 sides fenced, that is a square. In this case, with the length parallel to the wall, we have length=500/2 = 250 and width = 500/2/2 = 125 so the area is 250*125 = 31,250
This is greater than 166.67^2 = 27,779.
To solve this algebraically,
If x is the length parallel to the wall and y is the width,
x+2y=500
The area is
a = xy = (500-2y)y = 500y-2y^2
da/dy = 500-4y
da/dy = 0 when y=125
so x = 500-2y = 250
This is greater than 166.67^2 = 27,779.
To solve this algebraically,
If x is the length parallel to the wall and y is the width,
x+2y=500
The area is
a = xy = (500-2y)y = 500y-2y^2
da/dy = 500-4y
da/dy = 0 when y=125
so x = 500-2y = 250