kaplan 800 question

Hi,
Personnally, when manipulating inequatlities of negative numbers I convet them to inequalties of positive numbers to make them regular and easy to treat;

Fo this one,I would take the positive number (-x) which would be 1/3 < -x < 1/2

I would consider y^2 as the same of (-y)^2 and get,
1/25 < y^2 < 1/16 as 1/5 <-y < 1/4

then, the product -xy^2 as a positive number, would fit the following inequality : 1/75 < -xy^2 < 1/32

Now, it's easy to get xy^2 inequality as follows:

-1/32 < xy^2< -1/75


it's then obvious that the minimum value of xy^2 is -1/32

I hope I am right

For min and max values question, it is important to know the sign of
the end-result.

In this case, we know that the xy^2 will be negative. So, minimum
value of xy^2 will have to be a number which is farther away from zero.
(A simple example would be to compare -1 and -3, -3 is the minimum since it is farther away from 0 compared to -1.)

So, min (xy^2) is in essence min (x) * max (y^2) which yields
-1/2 * 1/16 = -1/32. Please note that we can split min (xy^2) as
above only because we already know what the sign of the
end result will be.

Oh, now i get it! thanks Stacey!!