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

Welcome, future engineer. Today, we are not just solving an algebra problem; we are exploring the topography of a cubic function. Imagine you are standing on a landscape defined by the function $f(x)=2x^3-9a{x}^2+12a^{2}x+1$.

This is a classic cubic curve that rises, falls, and rises again. Our mission is to find the specific value of the parameter $a$ that dictates the shape of this landscape, given that the peak (local maximum) occurs at $x=p$ and the valley (local minimum) occurs at $x=q$, with the elegant constraint $p^2=q$.

To understand where the peaks and valleys of our landscape lie, we must look at the slope. In calculus, the slope of the tangent line at any point is given by the first derivative, $f^{\prime }(x)$. We want to find the points where the landscape is perfectly flat—the critical points.

Let us differentiate our function:

Applying the power rule term by term, we obtain:

This quadratic expression is the key to our mystery. By setting $f^{\prime }(x)=0$, we find the exact locations where the slope is zero.

Setting the derivative to zero gives us:

To make this manageable, we divide the entire equation by $6$:

Now, we factorize. We are looking for two numbers that multiply to $2a^2$ and add to $-3a$. Those numbers are $-a$ and $-2a$. Thus, the equation becomes:

This reveals our two critical points: $x=a$ and $x=2a$. We must now determine which is the peak and which is the valley.

To distinguish between the maximum and the minimum, we invoke the Second Derivative Test. We differentiate $f^{\prime }(x)$ one more time:

Now, let us test our points. For $x=a$:

Since the problem states $a>0$, $-6a$ is negative. A negative second derivative indicates concavity downwards, which means we have found our local maximum. Thus, $p=a$.

For $x=2a$:

Since $a>0$, $6a$ is positive. A positive second derivative indicates concavity upwards, which means we have found our local minimum. Thus, $q=2a$.

We have our variables: $p=a$ and $q=2a$. The problem provides the final piece of the puzzle: $p^2=q$. Substituting our values, we get:

Rearranging this, we have $a^2-2a=0$, which factors to $a(a-2)=0$. This gives us two potential solutions: $a=0$ or $a=2$.

However, we must respect the constraint $a>0$. Therefore, we reject $a=0$ and accept $a=2$ as our definitive answer.