Divide the number 84 into two parts such that the product of one part and the square of other is maximum.

Options

A. 42, 42
B. 48, 36
C. 56, 28
D. None of These

Answer

To solve the problem of dividing the number 84 into two parts such that the product of one part and the square of the other part is maximized, we can follow these steps:

### Step 1: Define the Variables
Let one part be \( x \). Then the other part will be \( 84 - x \).

### Step 2: Define the Function to Maximize
We need to maximize the product of one part and the square of the other part. Therefore, we can express this as:
\[ P = x \cdot (84 - x)^2 \]

### Step 3: Expand the Function
Now, let's expand the function \( P \):
\[ P = x \cdot (84^2 - 168x + x^2) = x(7056 - 168x + x^2) \]
\[ P = 7056x - 168x^2 + x^3 \]

### Step 4: Differentiate the Function
Next, we differentiate \( P \) with respect to \( x \):
\[ \frac{dP}{dx} = 7056 - 336x + 3x^2 \]

### Step 5: Set the Derivative to Zero
To find the critical points, set the derivative equal to zero:
\[ 3x^2 - 336x + 7056 = 0 \]
Dividing the entire equation by 3:
\[ x^2 - 112x + 2352 = 0 \]

### Step 6: Solve the Quadratic Equation
Now we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -112 \), and \( c = 2352 \):
\[ x = \frac{112 \pm \sqrt{(-112)^2 - 4 \cdot 1 \cdot 2352}}{2 \cdot 1} \]
\[ x = \frac{112 \pm \sqrt{12544 - 9408}}{2} \]
\[ x = \frac{112 \pm \sqrt{3136}}{2} \]
\[ x = \frac{112 \pm 56}{2} \]
Calculating the two possible values:
1. \( x = \frac{168}{2} = 84 \)
2. \( x = \frac{56}{2} = 28 \)

### Step 7: Determine Which Value Maximizes the Product
We have two critical points: \( x = 84 \) and \( x = 28 \). However, \( x = 84 \) means the other part is 0, which does not make sense in this context. Thus, we take \( x = 28 \).

### Step 8: Find the Other Part
The other part is:
\[ 84 - x = 84 - 28 = 56 \]

### Conclusion
The two parts are \( 28 \) and \( 56 \).