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

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

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