Question 114823
The problem here is to compute the volume of a rectangular solid.  The dimensions of the solid are 15 feet by 15 feet by 1/2 foot.


We know that the volume of a rectangular solid is given by {{{V=lwh}}} where V is the volume, l is the length, w is the width, and h is the height.  So, in this case, the Volume is:


{{{V=15*15*(1/2)=225/2=112.5*ft^3}}}


You were given that there are 27 cubic feet in a cubic yard, so it is just a matter of dividing the number of cubic feet required, so:


{{{112.5/27=(4+(1/6))yd^3}}}


BUT you were told to round to the nearest cubic yard, so the 'correct' answer according to the parameters of the problem is {{{4*yd^3}}}.


However, have you noticed the problem with that already?  If you only have 4 yards delivered, you are going to end up with about a 3 foot by 3 foot square in one corner of the basketball court that doesn't have any concrete at all -- or perhaps an 18 square foot section that is only half as thick as it is supposed to be.  So, in practical terms, rather than rounding to the nearest cubic yard, you should be rounding UP to the next cubic yard.  In that case, the answer would be {{{5*yd^3}}}.  Another practical consideration for this problem is that even if you can find a concrete contractor who will deliver less than a full truckload (last time I looked that was about 6.5 yards), you are likely to pay a premium price per yard.  You might want to discuss these considerations with your instructor.


Super-Double-Plus-Extra-Credit:  How much of a real basketball court can you fit into a 15' X 15' area?