# Thread: Find the possible minimum

1. ## Find 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.)

2. Originally Posted by fcabanski
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.)
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
$\sqrt{1} = 1$
$\sqrt{4} = 2$
use 1 for the numerator, x=1

0<=y<=8.
$(2\times 0+1) = 1$
$(2\times 8+1) = 17$
use 17 for the denominator, y=8

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

3. Originally Posted by fcabanski
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.)
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 ?