Animated Solution for Mathematics - Differentiation: If the function $f(x)=2x^3-9a{x}^2+12a^{2}x+1$, where $a>0$, attains its local maximum and local minimum values at $p$ and $q$, respectively, such that $p^2=q$, then $f(3)$ is equal to:

Imagine you are standing on the graph of the function $f(x)=2x^3-9a{x}^2+12a^{2}x+1$. Because the leading coefficient is positive, this is a dynamic, sweeping path that climbs from negative infinity, reaches a peak, descends into a valley, and then surges upward again.

The points where this curve turns—the peak and the valley—are our local maximum and local minimum. We are told the maximum occurs at $p$ and the minimum at $q$, with the constraint that $p^2=q$.

Let us embark on the journey to find $f(3)$.

To find these turning points, we need to identify where the slope of the tangent line is perfectly horizontal. We calculate the first derivative, $f^{\prime }(x)$:

Applying the power rule, we obtain:

This derivative acts as our compass. To find the critical points, we set $f^{\prime }(x)=0$:

Dividing by $6$ simplifies our task immensely:

Next, we factorize this quadratic equation. We look for two numbers that multiply to $2a^2$ and add to $-3a$, which are $-a$ and $-2a$. Thus, the equation becomes:

This yields two critical points: $x=a$ and $x=2a$. Since we are given that $a>0$, we know for certain that $a<2a$.

For a cubic with a positive leading coefficient, the first critical point from the left is the local maximum, and the second is the local minimum. Therefore, $p=a$ and $q=2a$.

We are armed with the condition $p^2=q$. Substituting our values, we get:

Rearranging this, we have $a^2-2a=0$, which factors to $a(a-2)=0$. Since the problem explicitly states $a>0$, we must reject $a=0$.

Thus, $a=2$. With $a$ determined, our function is fully revealed:

Finally, we evaluate $f(3)$:

The journey from the abstract geometry of a cubic curve to the concrete value of $37$ is complete.