A value of $c$ for which the minimum value of $f(x) = x^2 - 4cx + 8c$ is greater than the maximum value of $g(x) = -x^2 + 3cx - 2c$, is

We have two functions:

$f(x) = x^2 - 4cx + 8c$ — a parabola opening upward, so it has a minimum.

$g(x) = -x^2 + 3cx - 2c$ — a parabola opening downward, so it has a maximum.

We need to find $c$ such that the minimum value of $f$ is greater than the maximum value of $g$.

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For an upward-opening quadratic $ax^2 + bx + k$, the minimum value is $k - \dfrac{b^2}{4a}$.

Here $a = 1,\ b = -4c,\ k = 8c$, so:

$\min f = 8c - \dfrac{(-4c)^2}{4(1)} = 8c - \dfrac{16c^2}{4} = 8c - 4c^2$

This comes from completing the square: $f(x) = (x - 2c)^2 + (8c - 4c^2)$. The squared part is always $\geq 0$, so the minimum value is $8c - 4c^2$.

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For a downward-opening quadratic, the maximum value is $k - \dfrac{b^2}{4a}$.

Here $a = -1,\ b = 3c,\ k = -2c$, so:

$\max g = -2c - \dfrac{(3c)^2}{4(-1)} = -2c + \dfrac{9c^2}{4}$

This comes from completing the square: $g(x) = -\left(x - \dfrac{3c}{2}\right)^2 + \left(-2c + \dfrac{9c^2}{4}\right)$. The squared part is always $\leq 0$, so the maximum value is $-2c + \dfrac{9c^2}{4}$.

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Now we apply the condition $\min f > \max g$:

$8c - 4c^2 > -2c + \dfrac{9c^2}{4}$

Bringing everything to one side:

$10c - 4c^2 - \dfrac{9c^2}{4} > 0$

Converting $4c^2 = \dfrac{16c^2}{4}$:

$10c - \dfrac{25c^2}{4} > 0$

Multiplying both sides by $4$:

$40c - 25c^2 > 0$

Factoring:

$5c(8 - 5c) > 0$

A product of two factors is positive when both share the same sign:

$c > 0$ and $8 - 5c > 0$ gives $0 < c < \dfrac{8}{5}$

$c < 0$ and $8 - 5c < 0$ gives $c < 0$ and $c > \dfrac{8}{5}$, which is impossible.

So the valid range is $0 < c < \dfrac{8}{5}$.

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Since $\dfrac{8}{5} = 1.6$, any value of $c$ strictly between $0$ and $1.6$ works.

Among the given options, $c = \dfrac{1}{2} = 0.5$ lies in this interval.

$\boxed{c = \dfrac{1}{2}}$