Posted by angel on Wednesday, April 18, 2012 at 8:44pm.
He will need one length and two widths
let the length by y ft
let the width be x ft
2x + y = 80
y = 80-2x
area = xy
= x(80-2x)
= -2x^2 + 80x
The x value to produce the maximum area is the x of the vertex of this parabola.
If you know Calculus
d(area)/dx = -4x + 80= 0 for a max of area
x = 20
then y = 80-2x = 40
the field should be 20 by 40 , with 40 being the one side.
non-Calculus:
the x of the vertex of y = ax^2 + bx +c is -b/(2a)
which in this case is -80/(2(-2)) = 20
continue as above.
or you can complete the square:
area = -2(x^2 - 40x + 400 - 400 )
= -2(x-20)^2 + 800
max of area is 800 when x = 20 and y = 40
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