If x is a positive real number, then the greatest value of `(7-x)(x+5)^(2)`, is

Options

A. 256
B. 128
C. 64
D. 16

Explanation

To find the greatest value of the expression \( (7-x)(x+5)^2 \) for positive real numbers \( x \), we can follow these steps:

### Step 1: Define the function
Let \( f(x) = (7-x)(x+5)^2 \).

### Step 2: Differentiate the function
To find the critical points, we need to differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}[(7-x)(x+5)^2] \]
Using the product rule, we have:
\[ f'(x) = (7-x) \cdot \frac{d}{dx}[(x+5)^2] + (x+5)^2 \cdot \frac{d}{dx}[7-x] \]
Calculating the derivatives:
\[ \frac{d}{dx}[(x+5)^2] = 2(x+5) \]
\[ \frac{d}{dx}[7-x] = -1 \]
Thus,
\[ f'(x) = (7-x)(2(x+5)) + (x+5)^2(-1) \]
Simplifying this gives:
\[ f'(x) = 2(7-x)(x+5) - (x+5)^2 \]

### Step 3: Set the derivative to zero
Now, we set \( f'(x) = 0 \):
\[ 2(7-x)(x+5) - (x+5)^2 = 0 \]
Factoring out \( (x+5) \):
\[ (x+5)[2(7-x) - (x+5)] = 0 \]
This gives us two cases to consider:
1. \( x + 5 = 0 \) (not valid since \( x \) is positive)
2. \( 2(7-x) - (x+5) = 0 \)

### Step 4: Solve the equation
Solving the second equation:
\[ 2(7-x) - (x+5) = 0 \]
Expanding this gives:
\[ 14 - 2x - x - 5 = 0 \]
Combining like terms:
\[ 14 - 5 - 3x = 0 \implies 9 = 3x \implies x = 3 \]

### Step 5: Evaluate the function at critical points
Now we evaluate \( f(x) \) at \( x = 3 \):
\[ f(3) = (7-3)(3+5)^2 = 4 \cdot 8^2 = 4 \cdot 64 = 256 \]

### Step 6: Check endpoints
Since \( x \) must be positive, we check the behavior as \( x \) approaches the boundaries:
- As \( x \to 0 \):
\[ f(0) = (7-0)(0+5)^2 = 7 \cdot 25 = 175 \]
- As \( x \to 7 \):
\[ f(7) = (7-7)(7+5)^2 = 0 \]

### Conclusion
The maximum value occurs at \( x = 3 \):
\[ \text{Greatest value of } (7-x)(x+5)^2 \text{ is } 256. \]