Let $$a > 0$$. If the function $$f(x) = 6x^3 - 45ax^2 + 108a^2x + 1$$ attains its local maximum and minimum values at the points $$x_1$$ and $$x_2$$ respectively such that $$x_1 x_2 = 54$$, then $$a + x_1 + x_2$$ is equal to :

The local maxima and minima of a differentiable function occur at the critical points where its first derivative is zero.

Given $$f(x)=6x^{3}-45ax^{2}+108a^{2}x+1$$ with $$a\gt 0$$.

First find the derivative:
$$f'(x)=\frac{d}{dx}\bigl(6x^{3}-45ax^{2}+108a^{2}x+1\bigr)$$
$$\;\;\;=18x^{2}-90ax+108a^{2}$$

Set the derivative equal to zero to locate the critical points:
$$18x^{2}-90ax+108a^{2}=0$$
Divide every term by $$18$$ to simplify:
$$x^{2}-5ax+6a^{2}=0 \; -(1)$$

The two roots of equation $$(1)$$ are the critical points. Let the roots be $$x_1$$ (local maximum) and $$x_2$$ (local minimum). For a quadratic $$x^{2}+bx+c=0$$, the relationships between its roots and coefficients are:
• Sum of roots $$x_1+x_2=-b$$
• Product of roots $$x_1x_2=c$$

Comparing with $$(1)$$, we get
$$x_1 + x_2 = 5a$$
$$x_1 x_2 = 6a^{2} \; -(2)$$

The question states that $$x_1 x_2 = 54$$. Using $$(2)$$:
$$6a^{2}=54$$
$$a^{2}=9$$
Because $$a\gt 0$$, take the positive root:
$$a=3$$

Now calculate $$x_1+x_2$$ using $$x_1+x_2=5a$$:
$$x_1+x_2 = 5(3)=15$$

Finally evaluate $$a + x_1 + x_2$$:
$$a + x_1 + x_2 = 3 + 15 = 18$$

Hence $$a + x_1 + x_2 = 18$$.

Option B (18)