Question 95706
START BY DRAWING A SKETCH OF THE 6 BY 10 PLOT. NOW DRAW A SMALL BORDER AROUND IT & LABEL THE WIDTH X.
THE AREA TO BE FILLED IN IS:
(6+2X)+(10+2X)-6*10
60+20X+12X+4X^2-60
4X^2+32X IS THE AREA TO BE CEMENTED TO A DEPTH OF 3 INCHES WHICH IS 3/12 FEET OR .25 FEET.
THE AMOUNT OF CEMENT IN FEET IS 38383=27 CUBIC FEET.
NEW WE SOLVE FOR THE VALUE OF X:
(4X^2+32X)*.25=27
X^2+8X-27=0
USING THE QUADRATIC EQUATION WE GET:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
X=(-8+-SQRT[8^8-4*1*-27])/281
X=(-8=-SQRT(64*108])/2
X=(-8+-SQRT172)/2
X=(-8+-13.11)/2
X=(-8+13.11)/2
X=5.11/2
X=2.56 FEET IS THE WIDTH OF THE BORDER NEEDED FOR THE 27 CUBIC FEET OF CEMENT.
FILL IN THESE MEASUREMENTS IN YOUR DIAGRAM AND VERIFY THE ANSWERS OR 2.56 FEET FOR THE WIDTH & .25 FEET FOR THE DEPTH.