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

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

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.

I think this is a high-quality question and I don't agree with the explanation. Isn't -1/100 lesser in value than -1/18? And -1/100 should be the answer as it could be obtained by using x=-1/5 and y=-1/4

I think this is a high-quality question and I don't agree with the explanation. Isn't -1/100 lesser in value than -1/18? And -1/100 should be the answer as it could be obtained by using x=-1/5 and y=-1/4

No. -1/100 > -1/18.

>> !!!

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I think this is a high-quality question and I don't agree with the explanation. Isn't -1/100 lesser in value than -1/18? And -1/100 should be the answer as it could be obtained by using x=-1/5 and y=-1/4

hi, 1/18>1/100, but the opposite is true if the negatives are taken.. -1/18<-1/100... put easier values to check say -1/1 and -1/2
_________________

Since the question is to find out the least possible value of an equation wherein two variables are in multiplication form, it's quite evident that we need to take the least possible values of x and y and put them into (x^2)*y in order to get the least possible value for the equation......

From the inequality, we are able to see that least value for x and y is (-1/3) and (-1/2) respectively.

So the answer is - ((-1/3)^2)*(-1/2) = (1/9)*(-1/2) = -1/18 ...So the answer is indeed "D"