I know that the formula for area of a rectangle is A=lw, where l and w are the length and width, respectively. I try to use other information in the problem to tell me something about l and w. I know that the formula for perimeter of a rectangle is P=2l+2w. The problem says that the perimeter is 22, so I write 22=2l+2w. Factoring out a 2 gives 22=2(l+w). I then divide by 2 to get 11=l+w. This tells me that the sum of l and w is 11. Since no other information about l and w is available, I must assume that they can be ANY two numbers that add up to 11. I need to find two numbers whose sum is 11 and whose product is one of the choices. I try whole numbers first because that's easier.
10+1=11; 10*1=10, but 10 is not a choice.
9+2=11; 9*2=18. Choice (d) is correct.
8+3=11; 8*3=24. Choice (b) is correct.
7+4=11; 7*4=28, but 28 is not a choice.
6+5=11; 6*5=30. Choice (a) is correct.
Now I only need to check choices (c) and (e). Checking choice (c), I write the following system of equations:
l+w=11; lw=121
I can solve the system by substitution. I solve the first equation for l and plug that into the other equation for l.
l=11-w
(11-w)w=121
Now, the second equation only has one variable, so I can solve it. I distribute the w on the left side to obtain 11w-w^2=121. Rearranging yields w^2-11w+121=0. I use the quadratic formula to find w and notice that the discriminant (the part under the radical) is negative. This means that the equation has no real solutions. In other words, there are NO real numbers l and w, whole or not, such that l+w=11 and lw=121.
I repeat the process for choice (e):
(11-w)w=108
11w-w^2=108
w^2-11w+108=0
I again obtain a negative discriminant, so there are no real solutions to the system l+w=11; lw=108, either. Therefore, only answers (a), (b), and (d) are correct.
