Bunuel wrote:

If \({-\frac{1}{3}} \le {x} \le {-\frac{1}{5}}\) and \({-\frac{1}{2}} \le {y} \le {-\frac{1}{4}}\), what is the least value of \(x^2*y\) possible?

A. \(-\frac{1}{100}\)

B. \(-\frac{1}{50}\)

C. \(-\frac{1}{36}\)

D. \(-\frac{1}{18}\)

E. \(-\frac{1}{6}\)

I got this question wrong in the test, because of a silly mistake, but now when I look at my mistake, I laugh. Anyways, here's the explanation

If the value has to be least, possible, the value must be negative and the value must have the highest modulus.

We know that the expression \(x^2*y\) will give a negative value for negative y, now our aim is to keep the modulus maximum.

so, \(x=1/3\) and \(y =-1/2\).

Thus the lowest value will be \(-1/18\)

Mistakenly, I calculated the value for the minimum modulus and got it wrong earlier.

Happy to learn...

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