If -1/2

Question

If  -\frac{1}{2} \leq x \leq -\frac{1}{3}  and  -\frac{1}{4} \leq y \leq -\frac{1}{5} , what is the minimum value of  x*y^2 ?

A.  -\frac{1}{75}

B.  -\frac{1}{50}

C.  -\frac{1}{48}

D.  -\frac{1}{32}

E.  -\frac{1}{16}

GMATinsight · Answer

For Minimum Value of x*y^2

1) The absolute value of x*y^2 must be Maximum and
2) Sign of x*y^2 must be negative

@x = -1/2 (Max absolute value), y = -1/4 (Max absolute value)

x*y^2 = -1/32

Answer: option D

ScottTargetTestPrep · Answer

We want the value of x to be as small as possible and the value of y^2 (which is positive) to be as large as possible.

The smallest value of x is -1/2, and the largest value of y^2 is (-1/4)^2 = 1/16.

Thus, the minimum value of x * y^2 is -1/2 x (-1/4)^2 = -1/2 x 1/16 = -1/32.

Answer: D

ishpreetanand · Answer

why are we taking the max value of y^2 and not the lowest value ?

dineshril · Answer

Hi Ispreet Sharing my thoughts .... On the number line larger the magnitude or larger the absolute value of a -ve number SMALLER is the number =>Eg -10 is smaller than -5 i.e -10<-5 BUT Mod of -10 is larger than -5 i.e |-10|>|-5| =>Eg \frac{-1}{2}=-0.5 < \frac{-1}{3}=-0.3 BUT | \frac{-1}{2}|=0.5 > | \frac{-1}{3}|=0.3 In the above Ques in discussion we have to find the MINIMUM Value of ( x*y^2 ) => Now ( x*y^2 ) will be -ve (since 'x' is -ve & y^2 is +ve) MINIMUM Value of ( x*y^2 ) can be obtained WHEN | x*y^2 | is MAXIMUM => Since 'x' is -ve we take the MINIMUM value of 'x' i.e x=\frac{-1}{2} AND => Since y^2 is +ve we take the MAXIMUM value of y^2 i.e y^2=(\frac{-1}{4})*(\frac{-1}{4})=\frac{1}{16} => The above combination gives MAXIMUM | x*y^2 | i.e MINIMUM ( x*y^2 ) Regards Dinesh

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If -1/2 <= x <= -1/3 and -1/4 <= y <= -1/5, what is the minimum value

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