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.
You would have to make 465=1550-7 1/2 x and find x Subtract 1550 from both sides -1085=-7 1/2x Divide -7 1/2 on both sides And the answer is 144 2/3 Sorry about ur bad luck
