Let M be the maximum value of the product of two positive integers when their sum is 66.

Question

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

Accepted Answer

$x+y=66$

$$
\begin{aligned}
& \frac{x+y}{2} \geq \sqrt{x y} \\
\Rightarrow & 33 \geq \sqrt{x y} \\
\Rightarrow & x y \leq 1089 \\
\therefore & M=1089 \\
S: & x(66-x) \geq \frac{5}{9} \cdot 1089 \\
& 66 x-x^{2} \geq 605 \\
\Rightarrow & x^{2}-66 x+605 \leq 0
\end{aligned}
$$

$\Rightarrow(x-55)(x-11) \leq 0 ; 11 \leq x \leq 55$

Therefore $S=\{11,12,13 \ldots 55\} $

$\Rightarrow n(S)=45$

Elements of $S$ which are multiple of 3 are

$$
\begin{aligned}
& 12+(n-1) 3=54 \Rightarrow 3(n-1)=42 \Rightarrow n=15 \\
& n(A)=15
\end{aligned}
$$

$\Rightarrow P(A)=\frac{15}{45}=\frac{1}{3}$

Solutions · 1