



 
Angelica, First, remember the formula P = 2L + 2W for the perimeter of a rectangle. Plugging in your amount of fence and simplifying gives you an expression for L in terms of W. Secondly, remember the formula A = LW for the area. You can use the expression you got last time to write this just in terms of W. Now if you are using calculus find the derivative of this and set it equal to 0. Any solution to this [here you should only find one] is a "critical point". The best solution must be at a critical point or somewhere where the area stops being defined (for instance when L=200 and W=0). Check all of these. If you have to do this without calculus, note that the function is quadratic so its graph is a parabola. By symmetry, the maximum of a parabola  if it has one, it might have a minimum instead  is halfway between its zeros  if it has any. In this case it obviously does: when L = 200, and when L=0. In both those cases the fence encloses no area at all. Good Hunting!
 


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 