Animated Solution for Mathematics - Differentiation: The maximum value of the function $f(x)=3x^3-18x^2+27x-40$ on the set $S=\{x\in R:x^2+30\le 11x\}$ is :

Welcome, fellow traveler of the JEE landscape! Today, we are embarking on a journey of optimization. We have a cubic function, $f(x)=3x^3-18x^2+27x-40$, confined to a specific playground, a set $S$ defined by the inequality $x^2+30\le 11x$.

Before we touch the function, we must understand our boundaries. The constraint $x^2+30\le 11x$ can be rewritten by bringing all terms to one side:

This is a classic quadratic inequality. We look for two numbers that multiply to $30$ and add to $-11$, which are $-5$ and $-6$. Thus, we factor the expression as:

Using the wavy curve method, we see that the product is negative or zero precisely between the roots $5$ and $6$. Our search space is the closed interval $x\in [5,6]$.

Now that we have our boundaries, we examine the function $f(x)=3x^3-18x^2+27x-40$. To find the maximum, we determine if this function is climbing or falling as we move from $5$ to $6$.

We find the derivative $f^{\prime }(x)$ to understand the slope:

Applying the power rule, we obtain:

Factoring out the common $9$, we get:

Here is where the intuition kicks in. We are only interested in the interval $x\in [5,6]$.

For any $x$ between $5$ and $6$, both $(x-1)$ and $(x-3)$ are clearly positive. Since both factors are positive, their product is positive, and thus $f^{\prime }(x)>0$ for all $x\in [5,6]$.

This tells us something profound: the function $f(x)$ is strictly increasing on our interval. It is a constant climb!

If a function is strictly increasing, it reaches its highest point at the very end of the interval. Therefore, the maximum value must occur at $x=6$.

We evaluate $f(6)$ as follows:

This simplifies to:

The $648$ terms cancel out perfectly. We are left with $162-40$, which equals $122$.

The maximum value of our function on the set $S$ is $122$.