If x, y are positive real numbers and 2x+3y=15, find the maximum value of x^{2}y^{3}.

We are tasked with finding the maximum value of the expression \( x^2 y^3 \) subject to the constraint \( 2x + 3y = 15 \), where \( x \) and \( y \) are positive real numbers.

### Step 1: Define the Lagrange multiplier function
We can solve this problem using the method of Lagrange multipliers. Define the function we want to maximize as:

\[
f(x, y) = x^2 y^3
\]

The constraint is \( 2x + 3y = 15 \). Define the constraint function as:

\[
g(x, y) = 2x + 3y - 15 = 0
\]

The Lagrange multiplier method tells us to solve the following system of equations:

\[
\nabla f = \lambda \nabla g
\]

Where \( \nabla f \) and \( \nabla g \) are the gradients of \( f \) and \( g \), respectively.

### Step 2: Compute the gradients
The gradient of \( f(x, y) = x^2 y^3 \) is:

\[
\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = \left( 2x y^3, 3x^2 y^2 \right)
\]

The gradient of \( g(x, y) = 2x + 3y - 15 \) is:

\[
\nabla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right) = (2, 3)
\]

### Step 3: Set up the system of equations
From the Lagrange multiplier method, we set:

\[
\nabla f = \lambda \nabla g
\]

This gives us the system:

1. \( 2x y^3 = \lambda \cdot 2 \)  (Equation 1)
2. \( 3x^2 y^2 = \lambda \cdot 3 \)  (Equation 2)
3. \( 2x + 3y = 15 \)  (Constraint)

### Step 4: Solve the system of equations

From Equation 1: \( 2x y^3 = 2 \lambda \), we can solve for \( \lambda \):

\[
\lambda = x y^3
\]

From Equation 2: \( 3x^2 y^2 = 3 \lambda \), substitute \( \lambda = x y^3 \) into this:

\[
3x^2 y^2 = 3(x y^3)
\]

Simplify:

\[
x^2 y^2 = x y^3
\]

Canceling one \( x \) and one \( y^2 \) (since \( x > 0 \) and \( y > 0 \)):

\[
x = y
\]

### Step 5: Substitute into the constraint
Now substitute \( x = y \) into the constraint \( 2x + 3y = 15 \):

\[
2x + 3x = 15
\]

\[
5x = 15
\]

\[
x = 3
\]

Since \( x = y \), we also have \( y = 3 \).

### Step 6: Find the maximum value of \( x^2 y^3 \)

Now that we have \( x = 3 \) and \( y = 3 \), substitute these values into the original expression \( x^2 y^3 \):

\[
x^2 y^3 = 3^2 \cdot 3^3 = 9 \cdot 27 = 243
\]

### Conclusion:
The maximum value of \( x^2 y^3 \) is \( \boxed{243} \).

Would you like further details or have any questions?

Here are five related questions:
1. What is the method of Lagrange multipliers and when is it used?
2. How do gradients help in solving optimization problems with constraints?
3. Can this problem be solved using other methods, such as substitution?
4. What would happen if the constraint were nonlinear, for example, \( 2x^2 + 3y^2 = 15 \)?
5. How do the conditions \( x > 0 \) and \( y > 0 \) influence the solution?

**Tip:** The method of Lagrange multipliers is particularly useful for optimizing functions subject to linear or nonlinear constraints, as it directly incorporates the constraint into the solution process.

If x, y are positive real numbers and 2x + 3y = 15, find the maximum value of x^2 y^3.

Explore how to find the maximum value of the expression x^2 y^3 subject to the constraint 2x + 3y = 15 using the method of Lagrange multipliers. This detailed solution involves gradient calculations, setting up the Lagrange system, and solving for optimal values. Suitable for undergraduate students studying calculus.

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