If x and y are integers such that 2

Question

If x and y are integers such that 2<x 8 and 2<y 9, what is the maximum value of 1/x-x/y ?

GMAT Panda · Accepted Answer

0 We use the constraints "2 < x 8" and "2 < y 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) = 1/x - x/y[[LB]]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) = 1/x - x/9 g(x+1) - g(x) = 1/x+1 - x+1/9 - (1/x - x/9) = -1/x(x+1) - 1/9 x = 3 We now evaluate at that pair and simplify the fraction before subtracting. f(3,9) = 1/3 - 3/9 3/9 = 3 13 3 = 1/3 1/3 - 1/3 = 0

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

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