Let a 0. If the function f(x) = 6x^3 - 45ax | vedclass

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(B) Given $f(x) = 6x^3 - 45ax^2 + 108a^2x + 1$. To find the local maximum and minimum,we find the derivative $f'(x)$: $f'(x) = 18x^2 - 90ax + 108a^2$.

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$18$

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(B) Given $f(x) = 6x^3 - 45ax^2 + 108a^2x + 1$. To find the local maximum and minimum,we find the derivative $f'(x)$: $f'(x) = 18x^2 - 90ax + 108a^2$. Setting $f'(x) = 0$ for critical points: $18(x^2 - 5ax + 6a^2) = 0$ $18(x - 2a)(x - 3a) = 0$. Thus,the critical points are $x = 2a$ and $x = 3a$. Since $f''(x) = 36x - 90a$,we check the nature of the points: $f''(2a) = 36(2a) - 90a = -18a $f''(3a) = 36(3a) - 90a = 18a > 0$ (Local minimum at $x_2 = 3a$). Given $x_1x_2 = 54$,we have $(2a)(3a) = 54$,so $6a^2 = 54$,which gives $a^2 = 9$. Since $a > 0$,we have $a = 3$. Then $x_1 = 2(3) = 6$ and $x_2 = 3(3) = 9$. Finally,$a + x_1 + x_2 = 3 + 6 + 9 = 18$.

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 $x_1x_2 = 54$,then $a + x_1 + x_2$ is equal to:

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