If the side parallel to the wall has length x meters, and the other side is y, we have
x+2y = 2000
The area a is just
a = xy = (2000-2y)y = 2000y - 2y^2
This is just a parabola, with its vertex (maximum value in this case) at y=500. So, x=1000, and the maximum area is 1000*500 m^2 or 1/2 km^2.
As always happens in this kind of problem, the maximum area for a given perimeter (or, equivalently, the minimum perimeter for a given area) is achieved when the available fence is equally divided between lengths and widths.
In this case, the length is x=1000 and there are two widths of y=500 each, or a total width allocation of 1000.
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? HELP PLS. show you work for understanding so i can do my home work.
1 answer