# Find the possible minimum

• June 9th 2009, 12:08 PM
fcabanski
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.)
• June 9th 2009, 12:29 PM
aidan
Quote:

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

• June 9th 2009, 12:32 PM
Amer
Quote:

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 ?