Let's assume
length of box is l
width of box is w
we know that
area of rectangle = length*width
[tex] 21=l*w [/tex]...................(1)
now, we are given
fencing length is 20 foot
fencing length will be equal to perimeter
and we know that
[tex] perimeter=2(l+w) [/tex]
so, we get
[tex] 2(l+w)=20 [/tex]
now, we can solve for l
[tex] l=10-w [/tex]
now, we can plug that in first equation
[tex] 21=(10-w)*w [/tex]
now, we can solve for w
[tex] 10w-w^2=21 [/tex]
[tex] w^2-10w+21=0 [/tex]
[tex] (w-7)(w-3)=0 [/tex]
[tex] w=7 , w=3 [/tex]
now, we can find length
[tex] l=10-3=7 [/tex]
or
[tex] l=10-7=3 [/tex]
so, option-B..................Answer
