If x and y are positive real numbers such that `x^(2)y^(3)=32` then the least value of `2x+3y` is

Options

A. 5
B. 10
C. 20
D. 15

Explanation

To find the least value of \(2x + 3y\) given that \(x^2 y^3 = 32\), we can use the method of Lagrange multipliers or apply the AM-GM inequality. Here, we will use the AM-GM inequality for simplicity.

### Step-by-Step Solution:

1. **Express the function to minimize**:
We need to minimize \(s = 2x + 3y\).

2. **Rewrite the expression**:
We can rewrite \(s\) as:
\[ s = 2x + y + y + y \]
This allows us to see that we have 5 terms: \(2x\) and three \(y\)s.

3. **Apply the AM-GM inequality**:
According to the AM-GM inequality:
\[ \frac{2x + y + y + y}{5} \geq \sqrt[5]{(2x)(y^3)} \]
Therefore, we have:
\[ 2x + 3y \geq 5 \sqrt[5]{(2x)(y^3)} \]

4. **Substitute the constraint**:
From the constraint \(x^2 y^3 = 32\), we can express \(y^3\) in terms of \(x\):
\[ y^3 = \frac{32}{x^2} \]
Now, substituting this into the AM-GM inequality gives:
\[ 2x + 3y \geq 5 \sqrt[5]{2x \cdot \frac{32}{x^2}} = 5 \sqrt[5]{\frac{64}{x}} = 5 \cdot \frac{4}{x^{2/5}} \]

5. **Minimize the expression**:
To minimize \(s\), we need to find the values of \(x\) and \(y\) such that \(x^2 y^3 = 32\) holds true. We can set \(x = y\) to simplify our calculations:
\[ x^2 x^3 = 32 \implies x^5 = 32 \implies x = 2 \]
Since \(x = y\), we also have \(y = 2\).

6. **Calculate the minimum value**:
Now substituting \(x = 2\) and \(y = 2\) back into \(s\):
\[ s = 2(2) + 3(2) = 4 + 6 = 10 \]

Thus, the least value of \(2x + 3y\) is \(\boxed{10}\).