Quadratics Extremum Comparison

For $f(x) = x^2 - 4cx + 8c$, the minimum occurs at $x = 2c$, giving: $\text{Min}(f) = (2c)^2 - 4c(2c) + 8c = -4c^2 + 8c$ For $g(x) = -x^2 + 3cx - 2c$, the maximum occurs at $x = \frac{3c}{2}$, giving: $\text{Max}(g) = -\left(\frac{3c}{2}\right)^2 + 3c\left(\frac{3c}{2}\right) - 2c = \frac{9c^2}{4} - 2c$ We are given $\text{Min}(f) > \text{Max}(g)$: $-4c^2 + 8c > \frac{9c^2}{4} - 2c \implies 10c > \frac{25c^2}{4} \implies c\left(10 - \frac{25c}{4}\right) > 0$ This gives $0 < c < \frac{40}{25} = 1.6$. Wait, re-calculating: $-4c^2 - \frac{9c^2}{4} + 10c > 0 \implies -\frac{25c^2}{4} + 10c > 0 \implies c \in (0, 1.6)$. Wait, checking $c = 2$ vs options: among given options, $c = 2$ gives $-16 + 16 = 0$ vs $9 - 4 = 5$ (not greater). Let's re-verify: if $c = 1/2$, $\text{Min}(f) = -4(1/4) + 4 = 3$, $\text{Max}(g) = 9/16 - 1 = -7/16$. Here $3 > -7/16$. Wait! In the options, option 4 is $1/2$. Thus $c = 1/2$ works!