Let M be the maximum value of the product of two positive integers when their sum is 66. Let the sample space `S ={X in Z:X(66-X) ge 5/9M}` and the event `A ={X in S : X` is a multiple of 3}. The P(A) is equal to

Answer

To solve the problem step by step, we will follow the reasoning provided in the transcript and clarify each step.

### Step 1: Determine the maximum product of two integers whose sum is 66.

Let the two integers be \( x \) and \( y \). Given that \( x + y = 66 \), we can express \( y \) as \( y = 66 - x \).

The product \( P \) of these two integers is given by:
\[ P = x(66 - x) = 66x - x^2 \]

To find the maximum value of this product, we can complete the square or use calculus. The product is a downward-opening parabola, and its maximum occurs at the vertex.

The vertex \( x \) of the parabola \( P = -x^2 + 66x \) can be found using the formula \( x = -\frac{b}{2a} \):
\[ x = -\frac{66}{2 \cdot (-1)} = 33 \]
Thus, the maximum product occurs when both integers are equal:
\[ M = P(33) = 33(66 - 33) = 33 \times 33 = 1089 \]

### Step 2: Define the sample space \( S \).

The sample space \( S \) is defined as:
\[ S = \{ x \in \mathbb{Z} : x(66 - x) \geq \frac{5}{9}M \} \]
Substituting \( M = 1089 \):
\[ S = \{ x \in \mathbb{Z} : x(66 - x) \geq \frac{5}{9} \times 1089 \} \]
Calculating \( \frac{5}{9} \times 1089 \):
\[ \frac{5}{9} \times 1089 = 605 \]
Thus, we need:
\[ x(66 - x) \geq 605 \]

### Step 3: Rearranging the inequality.

Rearranging gives:
\[ x^2 - 66x + 605 \leq 0 \]
We can solve this quadratic inequality by finding the roots of the equation:
\[ x^2 - 66x + 605 = 0 \]
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{66 \pm \sqrt{66^2 - 4 \cdot 1 \cdot 605}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 66^2 - 4 \cdot 605 = 4356 - 2420 = 1936 \]
Thus, the roots are:
\[ x = \frac{66 \pm 44}{2} \]
Calculating the two roots:
\[ x_1 = \frac{110}{2} = 55, \quad x_2 = \frac{22}{2} = 11 \]
The quadratic \( x^2 - 66x + 605 \) opens upwards, so it is less than or equal to zero between its roots:
\[ 11 \leq x \leq 55 \]
Thus, the sample space \( S \) is:
\[ S = \{ 11, 12, 13, \ldots, 55 \} \]

### Step 4: Determine the event \( A \).

The event \( A \) is defined as:
\[ A = \{ x \in S : x \text{ is a multiple of } 3 \} \]
The multiples of 3 between 11 and 55 are:
\[ 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54 \]
Counting these, we find there are 15 multiples of 3.

### Step 5: Calculate the probability \( P(A) \).

The total number of elements in \( S \) is:
\[ n(S) = 55 - 11 + 1 = 45 \]
Thus, the probability \( P(A) \) is given by:
\[ P(A) = \frac{n(A)}{n(S)} = \frac{15}{45} = \frac{1}{3} \]

### Final Answer:
The probability \( P(A) \) is \( \frac{1}{3} \).