Question

Let $M$ be the maximum value of the product of two positive integers when their sum is 66 Let the sample space $S =\left\{x \in Z : x(66-x) \geq \frac{5}{9} M\right\}$ and the event $A =\{x \in S : x$ is a multiple of 3$\}$ Then $P ( A )$ is equal to

• A. $\frac{15}{44}$
• B. $\frac{1}{3}$
• C. $\frac{7}{22}$
• D. $\frac{1}{5}$

Answer 1

The Correct Option is B

To solve this problem, we need to find the maximum product of two positive integers whose sum is 66, and then determine the probability of an event within a defined sample space.

Let the two integers be \(x\) and \(y\) such that \(x + y = 66\). Then, their product is given by:

\(P = x(66 - x) = 66x - x^2\)

This is a quadratic equation in terms of \(x\) and can be rewritten as:

\(P = -x^2 + 66x\)

To find the maximum value, we calculate the vertex of the parabola represented by this equation. The vertex for a quadratic equation \(ax^2 + bx + c\) is given by \(x = -\frac{b}{2a}\). Here, \(a = -1\) and \(b = 66\), hence:

\(x = -\frac{66}{2 \times (-1)} = 33\)

Thus, when \(x = 33\), \(y = 66 - x = 33\), giving \(66x - x^2 = 33 \times 33 = 1089\).

This value, \(M = 1089\), is the maximum product. Next, we define the sample space:

\(S = \{x \in \mathbb{Z} : x(66-x) \geq \frac{5}{9} \times 1089\}\)

Calculate \(\frac{5}{9} \times 1089 = 605\). Therefore, \(x(66-x) \geq 605\).

Solving the inequality:

\(66x - x^2 \geq 605\)

\(x^2 - 66x + 605 \leq 0\)

Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -66\), \(c = 605\), we get:

\(x = \frac{66 \pm \sqrt{66^2 - 4 \times 605}}{2}\)

\(x = \frac{66 \pm \sqrt{4356 - 2420}}{2}\)

\(x = \frac{66 \pm \sqrt{1936}}{2}\)

\(x = \frac{66 \pm 44}{2}\)

The solutions are \(x = 55\) and \(x = 11\). Thus, \(S = \{11, 12, ..., 54, 55\}\) (inclusive range between the roots).

Event \(A\) includes numbers that are multiples of 3:

\(A = \{12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54\}\)

The total number of outcomes in \(S\) is \(55 - 11 + 1 = 45\).

The number of favorable outcomes is 15. Therefore, the probability \(P(A) = \frac{15}{45} = \frac{1}{3}\).

Thus, the correct answer is \(\frac{1}{3}\).