Animated Solution for Mathematics - Differentiation: The maximum volume (in cu. m) of the right circular cone having slant height 3m is :

To solve this, we must first translate our physical intuition into the language of mathematics. We have a slant height $l=3$. Let the radius be $r$ and the height be $h$.

If we try to work with $r$ and $h$ directly, we get tangled in the constraint $r^2+h^2=l^2$. Instead, let us introduce the semi-vertical angle $\theta$. This angle is our master key, as it defines the 'opening' of the cone.

By using $\theta$, we can express both $r$ and $h$ in terms of a single variable. From the right-angled triangle formed by the slant height, radius, and vertical height, we see that:

Just like that, we have reduced a two-variable problem into a single-variable masterpiece.

Now, we recall the classic formula for the volume of a cone: $V=\frac{1}{3}\pi r^{2}h$. Substituting our expressions for $r$ and $h$, we get:

Let us pause and appreciate the structure here. Squaring the radius gives us $9{\sin }^2\theta$. When we multiply this by the height $3\cos \theta$ and the factor $\frac{1}{3}\pi$, the constants simplify beautifully.

We are left with the following function:

This is the function we must maximize. It is elegant, it is clean, and it is ready for the power of calculus.

To find the maximum, we must find where the slope of our volume function is zero. We need to calculate $\frac{dV}{d\theta }$. Using the product rule on $V=9\pi {\sin }^2\theta \cos \theta$, we differentiate:

Simplifying this, we get:

Setting this derivative to zero, we factor out $\sin \theta$:

Since $\sin \theta$ cannot be zero (that would mean no cone!), we must have $2{\cos }^2\theta ={\sin }^2\theta$. This leads us to the beautiful result: ${\tan }^2\theta =2$, or $\tan \theta =\sqrt{2}$.

We are almost there. We know $\tan \theta =\sqrt{2}$. Imagine a right triangle where the opposite side is $\sqrt{2}$ and the adjacent side is $1$.

By the Pythagorean theorem, the hypotenuse is $\sqrt{(\sqrt{2}{)}^2+1^2}=\sqrt{3}$. Thus, $\sin \theta =\frac{\sqrt{2}}{\sqrt{3}}$ and $\cos \theta =\frac{1}{\sqrt{3}}$.

Substituting these back into our volume expression:

The math simplifies perfectly:

There it is! The maximum volume is $2\sqrt{3}\pi$ cubic meters. You have successfully navigated the geometry, the trigonometry, and the calculus to find the peak.