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

For Detailed Answers:

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).

Answer: D.

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