If n is such that `36 le n le 72`, then `x = (n^(2) + 2sqrt(n)(n+4)+16)/(n+4sqrt(n) + 4)` satisfies

Options

A. `20 lt x lt 54`
B. `23 lt x lt 58`
C. `25 lt x lt 64`
D. `28 lt x lt 60`

Answer

To solve the problem, we need to evaluate the expression for \( x \) given the constraints on \( n \) and determine the range of values that \( x \) can take.

### Given:
\[ 36 \leq n \leq 72 \]
\[ x = \frac{n^2 + 2\sqrt{n}(n + 4) + 16}{n + 4\sqrt{n} + 4} \]

### Step 1: Substitute the lower bound \( n = 36 \)

First, we will calculate \( x \) when \( n = 36 \).

1. Calculate \( n^2 \):
\[ n^2 = 36^2 = 1296 \]

2. Calculate \( 2\sqrt{n}(n + 4) \):
\[ \sqrt{36} = 6 \quad \Rightarrow \quad 2\sqrt{36}(36 + 4) = 2 \cdot 6 \cdot 40 = 480 \]

3. Add \( 16 \):
\[ 1296 + 480 + 16 = 1792 \]

4. Calculate \( n + 4\sqrt{n} + 4 \):
\[ 4\sqrt{36} = 24 \quad \Rightarrow \quad 36 + 24 + 4 = 64 \]

5. Now, substitute these values into \( x \):
\[ x = \frac{1792}{64} = 28 \]

### Step 2: Substitute the upper bound \( n = 72 \)

Next, we will calculate \( x \) when \( n = 72 \).

1. Calculate \( n^2 \):
\[ n^2 = 72^2 = 5184 \]

2. Calculate \( 2\sqrt{n}(n + 4) \):
\[ \sqrt{72} = 6\sqrt{2} \quad \Rightarrow \quad 2\sqrt{72}(72 + 4) = 2 \cdot 6\sqrt{2} \cdot 76 = 912\sqrt{2} \]

(For approximation, we can use \( \sqrt{2} \approx 1.414 \)):
\[ 912 \cdot 1.414 \approx 1297.568 \]

3. Add \( 16 \):
\[ 5184 + 1297.568 + 16 \approx 6500.568 \]

4. Calculate \( n + 4\sqrt{n} + 4 \):
\[ 4\sqrt{72} = 24\sqrt{2} \quad \Rightarrow \quad 72 + 24\sqrt{2} + 4 \approx 76 + 33.94 \approx 109.94 \]

5. Now, substitute these values into \( x \):
\[ x \approx \frac{6500.568}{109.94} \approx 59.18 \]

### Conclusion

From our calculations:
- When \( n = 36 \), \( x \approx 28 \).
- When \( n = 72 \), \( x \approx 59.18 \).

Thus, the range of \( x \) is:
\[ 28 \leq x \leq 59.18 \]

### Final Answer
The value of \( x \) satisfies:
\[ 28 < x < 60 \]