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If x and y are integers such that 2 < x \leq 8 and 2 < y \leq 9, what is the maximum value of \frac{1}{x} - \frac{x}{y} ?

Explanation	Calculations	Help
$We use the constraints "2 < x \leq 8" and "2 < y \leq 9". We set up the expression and reason that, for any fixed positive value of x, increasing y makes the subtractive term smaller, which makes the whole expression larger. So we pick the largest allowed y.$	$f (x, y) = \frac{1}{x} - \frac{x}{y} y = 9$
$With y fixed at its largest value, we compare consecutive x values to see how the expression changes. The difference is negative, so the value goes down as x goes up. We then choose the smallest allowed x.$	$g (x) = f (x, 9) = \frac{1}{x} - \frac{x}{9} g (x + 1) - g (x) = \frac{1}{x + 1} - \frac{x + 1}{9} - (\frac{1}{x} - \frac{x}{9}) = - \frac{1}{x ( x + 1 )} - \frac{1}{9} x = 3$
$We now evaluate at that pair and simplify the fraction before subtracting.$	$f (3, 9) = \frac{1}{3} - \frac{3}{9} \frac{3}{9} = \frac{3 \times 1}{3 \times 3} = \frac{1}{3} \frac{1}{3} - \frac{1}{3} = 0$	Theory & Tactics Method Card FDPR2-A Click to view full details

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If x and y are integers such that 2 < x \leq 8 and 2 < y \leq 9, what is the maximum value of \frac{1}{x} - \frac{x}{y} ?

Explanation	Calculations	Help
$We use the constraints "2 < x \leq 8" and "2 < y \leq 9". We set up the expression and reason that, for any fixed positive value of x, increasing y makes the subtractive term smaller, which makes the whole expression larger. So we pick the largest allowed y.$	$f (x, y) = \frac{1}{x} - \frac{x}{y} y = 9$
$With y fixed at its largest value, we compare consecutive x values to see how the expression changes. The difference is negative, so the value goes down as x goes up. We then choose the smallest allowed x.$	$g (x) = f (x, 9) = \frac{1}{x} - \frac{x}{9} g (x + 1) - g (x) = \frac{1}{x + 1} - \frac{x + 1}{9} - (\frac{1}{x} - \frac{x}{9}) = - \frac{1}{x ( x + 1 )} - \frac{1}{9} x = 3$
$We now evaluate at that pair and simplify the fraction before subtracting.$	$f (3, 9) = \frac{1}{3} - \frac{3}{9} \frac{3}{9} = \frac{3 \times 1}{3 \times 3} = \frac{1}{3} \frac{1}{3} - \frac{1}{3} = 0$	Theory & Tactics Method Card FDPR2-A Click to view full details

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