What is the possible minimum for the expression with the range of x and y given as |SQRT X| / (2y +1) when x and y are given as:

1<=x<=4 0<=y<=8

?

(That's absolute value of the square root of x in the first part.)

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- Jun 9th 2009, 12:08 PMfcabanskiFind the possible minimum
What is the possible minimum for the expression with the range of x and y given as |SQRT X| / (2y +1) when x and y are given as:

1<=x<=4 0<=y<=8

?

(That's absolute value of the square root of x in the first part.) - Jun 9th 2009, 12:29 PMaidan
If you are looking for the minimum in a fraction,

then you want the smallest possible numerator and the largest possible denominator.

1<=x<=4

$\displaystyle \sqrt{1} = 1$

$\displaystyle \sqrt{4} = 2$

use 1 for the numerator, x=1

0<=y<=8.

$\displaystyle (2\times 0+1) = 1 $

$\displaystyle (2\times 8+1) = 17 $

use 17 for the denominator, y=8

$\displaystyle \frac {\sqrt{1}}{2\times 8+1} =\frac {1}{17} =$ the minimum

- Jun 9th 2009, 12:32 PMAmer
make denominator as large as possible and the numerator as small as possible

what is the max value for denominator in the interval that they gave you it is 1 or 4 ..or ??

and the numerator as small as possible it is 4 or 3 or ??

did recognize what I mean ?

someone reply before me - Jun 9th 2009, 12:43 PMfcabanski
Thanks...I knew I had to use the largest X and smallest Y, but I went blank on words for the explanation.