Question

The minimum value of \(\frac{3(6+x)(x+12)}{2(4+x)}\) where x>-4 is 

• A. 18
• B. 27
• C. 36
• D. 45

Answer 1

The Correct Option is B

To find the minimum value of the expression \( \frac{3(6+x)(x+12)}{2(4+x)} \) where \( x > -4 \), we need to simplify and analyze the expression.

1. First, let's expand the numerator of the given expression:

\[ 3(6+x)(x+12) = 3((6+x)(x+12)) = 3(x^2 + 12x + 6x + 72) = 3(x^2 + 18x + 72) \]

1. Now, distribute the 3:

\[ = 3x^2 + 54x + 216 \] 

1. The expression now becomes:

\[ \frac{3x^2 + 54x + 216}{2(4+x)} \]

1. To simplify further, we try to manipulate the expression in a way that could reveal its minimum value. Using calculus or by estimating critical points would be useful here.

Let's separate terms to better analyze them and find the minimum points. We consider two approaches: either studying domain intervals or direct derivation analysis for finding local minima (calculus). Here, we'll further simplify observationally with some trial values:

1. By trial values, let's substitute some integers:
   • When \( x = 0 \), 
     \[ \frac{3(6+0)(0+12)}{2(4+0)} = \frac{3 \cdot 6 \cdot 12}{2 \cdot 4} = \frac{216}{8} = 27 \]
2. For verification , calculate for others such as \( x = 1 \) or \( x = -2 \) might also help:
   • When \( x = 1 \), 
     \[ \frac{3(6+1)(1+12)}{2(4+1)} = \frac{3 \cdot 7 \cdot 13}{2 \cdot 5} = \frac{273}{10} = 27.3 \]
   • When \( x = -2\), 
     \[ \frac{3(6-2)(-2+12)}{2(4-2)} = \frac{3 \cdot 4 \cdot 10}{2 \cdot 2} = \frac{120}{4} = 30 \]
3. From these substitutions, the minimum value we obtained is \( \textbf{27} \) with \( x = 0 \). Thus, given the options, the correct and minimum value of the expression is \( \boxed{27} \).