What is the least value of x^2*y possible

If ${-\frac{1}{3}}\leq{x}\leq{-\frac{1}{5}}$ and ${-\frac{1}{2}}\leq{y}\leq{-\frac{1}{4}}$, what is the least value of $x^2*y$ possible?
A. -1/100
B. -1/50
C. -1/36
D. -1/18
E. -1/6

To get the least value of $x^2*y$, which obviously will be negative, try to maximize absolute value of $x^2*y$, as more is the absolute value of a negative number "more" negative it is (the smallest it is).
To maximize $|x^2*y|$ pick largest absolute values possible for $x$ and $y$: $(-\frac{1}{3})^2*(-\frac{1}{2})=-\frac{1}{18}$. Notice that: -1/18<-1/36<-1/50<-1/100, so -1/100 is the largest number and -1/18 is the smallest number (we cannot obtain -1/6 from x^2*y or else it would be the correct answer).