Question 143841This question is from textbook College Algebra
: PROBLEM:DIMENSIONS OF A PARKING LOT:
A SHOPPING CENTER HAS A RECTANGULAR AREA OF 40,000 YD^2 ENCLOSED ON THREE SIDES FOR A PARKING LOT. THE LENGTH IS 200 YD MORE THAN TWICE THE WIDTH. WHAT ARE THE DIMENSIONS OF THE LOT?
** IN THIS PRBLEM I ALWAYS SEEM TO KEEP GETTING A NEGATIVE ANSWER... I DO NOT UNDERSTAND THE PROPER WAY OF EVEN SOLVING THIS PROBLEM.. PLEASE HELP BY SHOWING ME HOW YOU WOULD WORK THIS THURALLY OUT.. STEP BY STEP, THANKS SO MUCH :)
This question is from textbook College Algebra
Answer by checkley77(12844)   (Show Source):
You can put this solution on YOUR website! LENGTH=2W+200
WIDTH=W
FORMULA FOR AREA IS:
AREA=L*W
40,000=(2W+200)W
40,000=2W^2+200W
2W^2+200W-40,000=0
2(W^2+100-20,000)=0
2(W-100)(W+200)=0
W-100=0
W=100 YDS. FOR THE WIDTH.
L=2*100+200
L=200+200
L=400
PROOF:
40,000=400*100
40,000=40,000
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