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.
I believe the correct answer is the second option. The correct despcription of the said figure would be a rectangle since from the choices it is the only shape where all angles are the same value which is 90. A kite can have different angles at each corner. A rhombus has two pairs of different angle measurements as well as a trapezoid.
