Animated Solution for Mathematics - Sequence and Series: Let $x,y>0$. If $x^3{y}^2=2^{15}$, then the least value of $3x+2y$ is

Mastering the AM-GM Dance

Welcome, future engineer. Today, we are not just solving an algebra problem; we are learning to see the hidden architecture of numbers.

When you look at a problem like this—minimizing a sum given a product constraint—you are looking at the heartbeat of competitive mathematics. It is a scenario that appears time and time again in the JEE Advanced, and once you master the intuition behind it, you will never fear it again.

We are given $x,y>0$ and the constraint $x^3{y}^2=2^{15}$. We want to minimize $3x+2y$.

Most students look at this and immediately try to substitute variables. They might try to express $y$ in terms of $x$, plug it into the sum, and then use calculus to find the minimum.

While that is a valid path, it is often the long, winding road. In the JEE, we value elegance. We value the 'surgical' approach. The Arithmetic Mean-Geometric Mean (AM-GM) inequality is that scalpel.

Why AM-GM? Because it is the bridge between addition and multiplication. It tells us that for any set of positive numbers, the average (Arithmetic Mean) is always greater than or equal to the product-based average (Geometric Mean).

Mathematically, for $n$ terms $a_1,a_2,\dots ,a_n$, we have:

Here is where the magic happens. We have $3x+2y$. If we just take the AM-GM of these two terms, we get $\frac{3x+2y}{2}\ge \sqrt{6xy}$.

This doesn't help us because our constraint is $x^3{y}^2$, not $xy$. We need the product of our terms to be $x^3{y}^2$.

This is the 'Aha!' moment. We need to split our terms so that when we multiply them, the exponents align perfectly with our constraint. We have $x^3$, so we need three $x$'s. We have $y^2$, so we need two $y$'s.

By splitting $3x$ into $x+x+x$ and $2y$ into $y+y$, we create a set of five terms: $x,x,x,y,y$.

Now, look at the beauty of this. The sum of these five terms is exactly $3x+2y$. And their product? It is $x\cdot x\cdot x\cdot y\cdot y=x^3{y}^2$. We have perfectly aligned our objective with our constraint.

Let us apply the inequality to our five terms. The Arithmetic Mean is:

And the Geometric Mean is:

By the AM-GM inequality, we know that $AM\ge GM$. Therefore:

Now, we substitute the given value $x^3{y}^2=2^{15}$ into our inequality:

This is where the fear of large exponents should vanish. Remember your laws of indices: $(a^m{)}^n=a^{m\cdot n}$.

Here, the fifth root is the same as raising to the power of $\frac{1}{5}$. So, $(2^{15}{)}^{\frac{1}{5}}=2^{15\cdot \frac{1}{5}}=2^3$. And $2^3$ is simply $8$.

Our inequality simplifies to:

We are almost there. To isolate our target expression, $3x+2y$, we multiply both sides by $5$:

And there it is. The minimum value of $3x+2y$ is $40$.

Think about what we just did. We didn't need complex derivatives or messy substitutions. We used the fundamental symmetry of numbers.

We transformed a constraint into a boundary, and in doing so, we found the floor of the function. This is the power of the JEE mindset—not brute force, but strategic, elegant, and precise application of mathematical laws. Keep this 'splitting' technique in your arsenal; it is the key to unlocking many such problems in your journey ahead.