Find minimum value of `(9-x) (2-x)`

Answer

To find the minimum value of the expression \( (9-x)(2-x) \), we can follow these steps:

### Step 1: Expand the expression
First, we will expand the expression \( (9-x)(2-x) \).

\[ (9-x)(2-x) = 9 \cdot 2 - 9x - 2x + x^2 \]
\[ = 18 - 11x + x^2 \]

### Step 2: Rewrite in standard quadratic form
We can rewrite the expression in standard quadratic form:

\[ x^2 - 11x + 18 \]

### Step 3: Find the vertex of the quadratic
To find the minimum value of a quadratic function \( ax^2 + bx + c \), we can use the vertex formula \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = -11 \).

\[ x = -\frac{-11}{2 \cdot 1} = \frac{11}{2} \]

### Step 4: Substitute \( x \) back into the expression
Now, we substitute \( x = \frac{11}{2} \) back into the quadratic expression to find the minimum value.

\[ \text{Minimum value} = \left(\frac{11}{2}\right)^2 - 11\left(\frac{11}{2}\right) + 18 \]
Calculating each term:

1. \( \left(\frac{11}{2}\right)^2 = \frac{121}{4} \)
2. \( -11\left(\frac{11}{2}\right) = -\frac{121}{2} = -\frac{242}{4} \)
3. \( 18 = \frac{72}{4} \)

Now combine these:

\[ \text{Minimum value} = \frac{121}{4} - \frac{242}{4} + \frac{72}{4} \]
\[ = \frac{121 - 242 + 72}{4} = \frac{-49}{4} \]

### Final Result
Thus, the minimum value of \( (9-x)(2-x) \) is

\[ \boxed{-\frac{49}{4}} \]

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