Question

Please help me on this problem

Answer 1

If Mr. Sanchez has 16 feet of fencing, the perimeter of his fencing must be 16 feet too.  The perimeter is the sum of all the side lengths; since he has 16 feet of fencing, this is all he can use.

He wants the fencing to be a rectangle.  The area of a rectangle is equal to length times width: A=lw.  The perimeter of a rectangle is equal to the length plus width plus length plus width, or twice the length plus twice the width: P=2l+2w.

We know that the perimeter (P) must equal 16.  We can fill this in: 16=2l+2w.  What are some possible lengths and widths that would fulfill this?  What if l=1?  Then we can fill that in: 16=(2*1)+2w.  2*1=2.  So, 16=2+2w.  Subtract 2 from both sides to get 14=2w.  Then divide both sides by 2: w=7.  This is a possible fencing scenario: the length is 1, and the width is 7.  BUT, we need to get the greatest area possible.  The area of this rectangle is A=lw, which would be A=1*7.  The area of this fenced area would be 7 feet squared.

What if l=2?  Use a similar method.  16=(2*2)+2w.  2*2=4.  16=4+2w.  12=2w.  w=6.  A=lw.  A=2*6=12 ft squared.

l=3: 16=(2*3)+2w.  2*3=10.  16=6+2w.  10=2w.  w=5.  A=3*5=15 sq. ft.

l=4: 16=(2*4)+2w.  2*4=8.  16=8+2w.  8=2w.  w=4.  A=4*4=16 sq ft.

After this, it starts to repeat: if l=5, then w=3, and you get the same area as l=3: 15.  If l=6, w=2, and A=12.  If l=7, w=1, and A=7.  l=8 or w=8 is impossible, because then the other dimension is 0 and you get an area of 0.  You can't have any dimensions above 8, either, or you will get a perimeter greater than 16--and then your other dimension would be negative, which is impossible in length.

So the possible areas are 7, 12, 15, and 16.  What's the greatest area? 16.