Find two numbers whose sum is 15 and the square of one multiplied by the cube of the other is maximum.

Answer

To solve the problem of finding two numbers whose sum is 15 and the product of the square of one number and the cube of the other is maximized, we can follow these steps:

### Step 1: Define the Variables
Let the two numbers be \( x \) and \( y \). According to the problem, we know:
\[ x + y = 15 \]
From this, we can express \( y \) in terms of \( x \):
\[ y = 15 - x \]

### Step 2: Define the Function to Maximize
We need to maximize the expression given in the problem, which is the square of one number multiplied by the cube of the other. We can define the function \( f(x) \) as:
\[ f(x) = x^2 \cdot y^3 = x^2 \cdot (15 - x)^3 \]

### Step 3: Differentiate the Function
To find the maximum value, we first differentiate \( f(x) \) with respect to \( x \):
\[ f(x) = x^2 (15 - x)^3 \]
Using the product rule:
\[ f'(x) = 2x(15 - x)^3 + x^2 \cdot 3(15 - x)^2 \cdot (-1) \]
This simplifies to:
\[ f'(x) = 2x(15 - x)^3 - 3x^2(15 - x)^2 \]

### Step 4: Set the Derivative to Zero
To find the critical points, we set \( f'(x) = 0 \):
\[ 2x(15 - x)^3 - 3x^2(15 - x)^2 = 0 \]
Factoring out common terms:
\[ x(15 - x)^2 (2(15 - x) - 3x) = 0 \]
This gives us:
\[ x(15 - x)^2 (30 - 5x) = 0 \]
The solutions are:
1. \( x = 0 \)
2. \( 15 - x = 0 \) (which gives \( x = 15 \))
3. \( 30 - 5x = 0 \) (which gives \( x = 6 \))

### Step 5: Evaluate the Critical Points
We discard \( x = 0 \) and \( x = 15 \) since they do not yield valid pairs of numbers (the other number would be zero). Thus, we only consider \( x = 6 \).

### Step 6: Find the Corresponding Value of \( y \)
Using \( x = 6 \):
\[ y = 15 - x = 15 - 6 = 9 \]

### Step 7: Conclusion
The two numbers that satisfy the conditions of the problem are:
\[ \boxed{6 \text{ and } 9} \]