Find two numbers whose sum is 12 and the product of the square of one part and the 4th power of other is maximum.

Answer

To solve the problem of finding two numbers whose sum is 12 and the product of the square of one number and the fourth power 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 have:
\[ x + y = 12 \]

### Step 2: Express One Variable in Terms of the Other
From the equation \( x + y = 12 \), we can express \( y \) in terms of \( x \):
\[ y = 12 - x \]

### Step 3: Define the Product Function
We need to maximize the product \( P \) defined as:
\[ P = x^2 \cdot y^4 \]
Substituting \( y \) from Step 2 into the product function gives:
\[ P = x^2 \cdot (12 - x)^4 \]

### Step 4: Differentiate the Product Function
To find the maximum value of \( P \), we need to differentiate it with respect to \( x \):
\[ P' = \frac{d}{dx}(x^2 \cdot (12 - x)^4) \]
Using the product rule:
\[ P' = (x^2)'(12 - x)^4 + x^2((12 - x)^4)' \]
Calculating the derivatives:
\[ P' = 2x(12 - x)^4 + x^2 \cdot 4(12 - x)^3 \cdot (-1) \]
Simplifying gives:
\[ P' = 2x(12 - x)^4 - 4x^2(12 - x)^3 \]

### Step 5: Set the Derivative to Zero
To find the critical points, set \( P' = 0 \):
\[ 2x(12 - x)^4 - 4x^2(12 - x)^3 = 0 \]
Factoring out common terms:
\[ 2x(12 - x)^3 \left( (12 - x) - 2x \right) = 0 \]
This simplifies to:
\[ 2x(12 - x)^3(12 - 3x) = 0 \]

### Step 6: Solve for Critical Points
Setting each factor to zero gives:
1. \( 2x = 0 \) → \( x = 0 \)
2. \( 12 - x = 0 \) → \( x = 12 \)
3. \( 12 - 3x = 0 \) → \( x = 4 \)

### Step 7: Determine Maximum or Minimum
To determine whether these critical points are maxima or minima, we can use the second derivative test or evaluate the first derivative around these points.

### Step 8: Evaluate the Product at Critical Points
1. For \( x = 0 \): \( y = 12 \) → \( P = 0^2 \cdot 12^4 = 0 \)
2. For \( x = 12 \): \( y = 0 \) → \( P = 12^2 \cdot 0^4 = 0 \)
3. For \( x = 4 \): \( y = 8 \) → \( P = 4^2 \cdot 8^4 = 16 \cdot 4096 = 65536 \)

### Conclusion
The maximum product occurs at \( x = 4 \) and \( y = 8 \).

### Final Answer
The two numbers are \( 4 \) and \( 8 \).

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