2x+y = 2000
a = xy = x(2000-2x)
da/dx = 2000-4x
da/dx=0 when x=500
So, the maximum area is 500*1000 m^2
Note that this is achieved when the available fence is equally divided between lengths and widths.
A car dealership fences in a rectangular
area behind their building to secure
unsold vehicles. One length will be
the back wall of the dealership. What is
the maximum parking area that can
be created if they have 2 km of fencing
to use?
1 answer
1 answer