GMAT Practice Question: If x and y are integers such that 2<x \leq 8 and 2<y \leq 9...
Question
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} ?
- -3 \frac{1}{8}
- 0
- \frac{1}{4}
- \frac{5}{18}
- 2
Topics: fractions, max/min of function, inequalities
Solution
Step 1
We use the constraints \textbf{"}2 < x \leq 8\textbf{"} and \textbf{"}2 < y \leq 9\textbf{"}. 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}[[LB]]y = 9
Step 2
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
Step 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{\cancel{3} \times 1}{\cancel{3} \times 3} = \frac{1}{3} \frac{1}{3} - \frac{1}{3} = 0
Answer
0
