SOLUTION:
We are to choose between a Cheez-it's that cover a rectangle with;
(i) A length of 9 and a perimeter of 22,
(ii) A length of 5 and a perimeter of 20.
The prefer one would be the one with a greater area.
[tex]\begin{gathered} \text{Length = 9 units} \\ \text{Perimeter = 22 units} \\ P\text{ = 2(L + W)}=22\text{ } \\ 2(9\text{ + w) = 22} \\ \frac{2(9+w)}{2}=\text{ }\frac{22}{2} \\ 9+w\text{ = 11} \\ w\text{ = 11-9} \\ w\text{ = 2 units} \end{gathered}[/tex]
The area of this rectangle is;
[tex]\begin{gathered} A=LXW\text{ } \\ A\text{ =9X2} \\ A=18unit^2 \end{gathered}[/tex]
[tex]\begin{gathered} \text{Length }=\text{ 5 units} \\ \text{Perimeter = 20 units} \\ P\text{ = 2(L + W) = 20} \\ 2(5\text{ + w) = 20} \\ \frac{2(5+w)}{2}=\text{ }\frac{20}{2} \\ 5\text{ + w = }10 \\ w=10-5 \\ w\text{ = 5 units} \end{gathered}[/tex]
The area of this rectangle is;
[tex]\begin{gathered} A\text{ = L x W} \\ A\text{ = 5 x 5} \\ A=25units^2 \end{gathered}[/tex]
CONCLUSION:
Since the area of a rectangle of length of 5 and a perimeter of 20 is greater than that of a length of 9 and a perimeter of 22.
The preferred is the Cheez-it's that covers a rectangle of a length of 5 and a perimeter of 22.
