Question

A farmer wants to enclose a rectangular field along a river on three sides. If 4,800 feet of fencing is to be used, what dimensions will maximize the enclosed area?

Answer 1

notice the picture below

thus   $\bf Area =lw\implies A(w)=(4800-2w)w\implies A(w)=4800w-2w^2$

get the derivative of it, zero it out to get the critical points, and do a first-derivative test on them for any maxima

you'll only get only critical point anyway

Answer 2

L=length w=width
Area=L*w
Perimeter= 2w+L

4,800 = 2w+L
4,800-L = 2w
(4,800-L)/2 = w

A=2*((4,8000-L)/2) * L
A= (4800-L)*L
A= 4800L-L^2
A'= 4800-2L, maximize
0=4800-2L
2L=4800
L=2400, plug back in to first formula

w=(4,800-L)/2
w= (4,800-2400)/2
w=1200

To maximize the area, the width must be 1200ft and the length 2400