Divide 4 into two positive numbers such that the sum of the square of one and cube of the other is a minimum.

Answer

To solve the problem of dividing 4 into two positive numbers such that the sum of the square of one and the cube of the other is minimized, we can follow these steps:

### Step 1: Define the Variables
Let one of the numbers be \( x \). Then, the other number will be \( 4 - x \) since the total must equal 4.

### Step 2: Set Up the Function to Minimize
We need to minimize the function:
\[ f(x) = x^2 + (4 - x)^3 \]
This function represents the sum of the square of the first number and the cube of the second number.

### Step 3: Differentiate the Function
To find the minimum, we first differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}((4 - x)^3) \]
Using the power rule and chain rule:
\[ f'(x) = 2x + 3(4 - x)^2 \cdot (-1) \]
This simplifies to:
\[ f'(x) = 2x - 3(4 - x)^2 \]

### Step 4: Set the Derivative to Zero
To find critical points, set \( f'(x) = 0 \):
\[ 2x - 3(4 - x)^2 = 0 \]
Rearranging gives:
\[ 2x = 3(4 - x)^2 \]

### Step 5: Expand and Rearrange the Equation
Expanding the right side:
\[ 2x = 3(16 - 8x + x^2) \]
This leads to:
\[ 2x = 48 - 24x + 3x^2 \]
Rearranging gives:
\[ 3x^2 - 26x + 48 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3 \), \( b = -26 \), \( c = 48 \):
\[ x = \frac{26 \pm \sqrt{(-26)^2 - 4 \cdot 3 \cdot 48}}{2 \cdot 3} \]
Calculating the discriminant:
\[ x = \frac{26 \pm \sqrt{676 - 576}}{6} = \frac{26 \pm \sqrt{100}}{6} = \frac{26 \pm 10}{6} \]
This gives two potential solutions:
\[ x = \frac{36}{6} = 6 \quad \text{and} \quad x = \frac{16}{6} = \frac{8}{3} \]

### Step 7: Check the Validity of Solutions
Since both numbers must be positive and \( x = 6 \) would lead to \( 4 - 6 = -2 \), we discard \( x = 6 \). Thus, we take:
\[ x = \frac{8}{3} \]
The other number is:
\[ 4 - x = 4 - \frac{8}{3} = \frac{12 - 8}{3} = \frac{4}{3} \]

### Step 8: Verify Minimum
To confirm that this is a minimum, we can check the second derivative:
\[ f''(x) = 2 - 6(4 - x) \]
Evaluating at \( x = \frac{8}{3} \):
\[ f''\left(\frac{8}{3}\right) = 2 - 6\left(4 - \frac{8}{3}\right) = 2 - 6\left(\frac{12 - 8}{3}\right) = 2 - 6 \cdot \frac{4}{3} = 2 - 8 = -6 \]
Since \( f''(x) > 0 \), it confirms a local minimum.

### Final Answer
The two positive numbers that divide 4 such that the sum of the square of one and the cube of the other is minimized are:
\[ \frac{8}{3} \quad \text{and} \quad \frac{4}{3} \]