Please Help Math Question?
2016-05-02 1:42 pm
Ethan had 60m of fence with which to build a rectangular pet run. What is the maximum area he can enclose.
回答 (1)
2016-05-02 2:00 pm
Let x be the length and y be the width of the fence.
Perimeter = 2x+2y = 60
2y = 60-2x
y = 30-x
Area = xy
Area = x(30-x)
Area = 30x - x^2
dA/dx = 30 -2x = 0
2x=30
x= 15 (length)
y=30-15 = 15 (width)
Maximum area = xy = (15)(15) = 225 m^2
d^2A/dx^2 = -2 < 0 so the area is maximum when length =15 and width =15
Perimeter = 2x+2y = 60
2y = 60-2x
y = 30-x
Area = xy
Area = x(30-x)
Area = 30x - x^2
dA/dx = 30 -2x = 0
2x=30
x= 15 (length)
y=30-15 = 15 (width)
Maximum area = xy = (15)(15) = 225 m^2
d^2A/dx^2 = -2 < 0 so the area is maximum when length =15 and width =15
收錄日期: 2021-05-01 20:40:33
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