ishpreetanand wrote:
ScottTargetTestPrep wrote:
Bunuel wrote:
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. \(-\frac{1}{16}\)
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
why are we taking the max value of y^2 and not the lowest value ?
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