Question 320471
YOUR RECTANGLAR AREA WILL HAVE 2 EQUAL SIDES & 1 LENGTH.
P=2W+L
800=2W+L
L=800-2W
A=LW
15,200=(800-2W)W
15,200=800W-2W^2
2W^2-800W+15,200=0
2(W^2-400W+7,600)=0
2(W-380)(W-20)=0
W-380=0
W=380 FT. 
L=800-2*380=800-760=40 FT.
OR:
W-20=0
W=20 FT.
------------------------
L=800-2W
AREA=LW
(800-2W)*W
800W-2W^2
-2W^2+800W
-2(W^2-400W) NOW COMPLETE THE SQUARE.
(400/2)^2=200^2=40,000
-2(W^2-400W+40,000)=40,000
-2(W-200)^2=40,000
W-200=0
W=200 ANS. FOR EACH OF THE SHORT WIDTHS.
L=800-2*200
L=800-400=400 ANS. FOR THE LENGTH,
PROOF:
A=400*200=80,000 FT^2 IS THE MAX. AREA.
TEST:
USE 199 & 402 FT.
AREA=199*402=79,998 FT^2
TEST 201 & 398
AREA=201*398=79,998 FT^2