1. **State the problem:** We want to find the number $x$ that maximizes the difference between the number and its square, i.e., maximize the function $$f(x) = x - x^2.$$ 2. **Formula and rules:** The function is a quadratic function $$f(x) = x - x^2 = -x^2 + x.$$ Since the coefficient of $x^2$ is negative, the parabola opens downward, so it has a maximum point at its vertex. 3. **Find the vertex:** The vertex of a parabola $ax^2 + bx + c$ is at $$x = -\frac{b}{2a}.$$ Here, $a = -1$ and $b = 1$, so $$x = -\frac{1}{2 \times (-1)} = \frac{1}{2}.$$ 4. **Evaluate the function at $x=\frac{1}{2}$:** $$f\left(\frac{1}{2}\right) = \frac{1}{2} - \left(\frac{1}{2}\right)^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}.$$ 5. **Check the options:** The maximum occurs at $x=\frac{1}{2}$, which is not among the options exactly. The closest option is $\frac{3}{4}$ or $\frac{1}{5}$. Let's check $f(x)$ at these points: - At $x=\frac{1}{5}$: $$f\left(\frac{1}{5}\right) = \frac{1}{5} - \left(\frac{1}{5}\right)^2 = \frac{1}{5} - \frac{1}{25} = \frac{4}{25} = 0.16.$$ - At $x=\frac{3}{4}$: $$f\left(\frac{3}{4}\right) = \frac{3}{4} - \left(\frac{3}{4}\right)^2 = \frac{3}{4} - \frac{9}{16} = \frac{12}{16} - \frac{9}{16} = \frac{3}{16} = 0.1875.$$ 6. **Compare values:** - $f(\frac{1}{2}) = 0.25$ (maximum) - $f(\frac{3}{4}) = 0.1875$ - $f(\frac{1}{5}) = 0.16$ Since $\frac{1}{2}$ is not an option, the closest maximum among the options is $\frac{3}{4}$. **Final answer:** C. $\frac{3}{4}$