Let the perimeter be 2w + 280-2w = 280
A = w * (280-2w)
A = 280w - 2w^2
dA/dw = 280 - 4w
A is max when dA/dw = 0
Thus, w=70
The dimensions are 70x140 and the area is 9800
If you have 280 meters of fencing and want to enclose a rectangular area up against a long, straight wall, what is the largest area you can enclose?
2 answers
Area A=x.y
The perimeter of a rectangle is 2x+y= 280 solving for y give you,m y=280-2x, y is greater to zero.
A=x*y substituting y in the area equation give you
A=x(280-2x.)=280x-2x^2
The area is maximized when A'=0
so first find the derivative of A
A'=280-4x. A=0 when 280-4x=0 solve for x. x=280/4=70 so x=70
you subtitue x into the y equation which is y=280-2x so y = 280-2(70)=280-140=140
so y=140
Finally the dimensions are x=70, and y=140 so the area is 70*140=9800m^2
The perimeter of a rectangle is 2x+y= 280 solving for y give you,m y=280-2x, y is greater to zero.
A=x*y substituting y in the area equation give you
A=x(280-2x.)=280x-2x^2
The area is maximized when A'=0
so first find the derivative of A
A'=280-4x. A=0 when 280-4x=0 solve for x. x=280/4=70 so x=70
you subtitue x into the y equation which is y=280-2x so y = 280-2(70)=280-140=140
so y=140
Finally the dimensions are x=70, and y=140 so the area is 70*140=9800m^2
2 answers