Question 154768This question is from textbook
: For security a company will enclose a rectangle area of 11,200 ft^2 in the rear of its plant. One side will be bound by the building and the other three sides by fencing .If 300ft of fencing will be used, what will be the dimensions of the rectangle This question is from textbook
: For security a company will enclose a rectangle area of 11,200 ft^2 in the rear of its plant. One side will be bound by the building and the other three sides by fencing .If 300ft of fencing will be used, what will be the dimensions of the rectangle
Answer by checkley77(3624)  (Show Source):
You can put this solution on YOUR website!XY=11,200
2X+Y=300 OR Y=300-2X
X(300-2X)=11,200
300X-2X^2=11,200
2X^2-300X+11,200=0
2(X^2-150X+5,600)=0
2(X-70)(X-80)=0
X-70=0
X=70 FOR THE TWO SHORT SIDE
2*70+Y=300
140+Y=300
Y=300-140
Y=160 THE SINGLE LONG SIDE.
PROOF:
70*160=11,200
11.200
OR:
X-80=0
X=80 FOR THE TWO SHORT SIDES.
2*80+Y=300
160=Y=300
Y=300-160
X=140 FOR THE SINGLE LONG SIDE.
PROOF:
80*140=11,200
11,200=11,200
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