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

Q: Wait, why is the maximum product at $x=33$? Can't it be any other value?

This is a classic application of the AM-GM inequality, which states that for a fixed sum, the product of two numbers is maximized when they are equal. Since $x + y = 66$, the product $xy$ is maximized when $x = y = 33$. If you try any other pair, like $32$ and $34$, the product is $1088$, which is less than $33 \times 33 = 1089$.

Q: Why is the range $11 \leq x \leq 55$ and not $x \geq 11$ or $x \leq 55$?

Great question! The inequality is $x^2 - 66x + 605 \leq 0$. Since the coefficient of $x^2$ is positive, the parabola opens upwards, but we are looking for the region where the quadratic is *less than or equal to* zero. This region lies *between* the two roots of the equation $x^2 - 66x + 605 = 0$. If the inequality were $\geq 0$, the solution would be the regions outside the roots.

Q: Why do we add 1 when counting the number of integers in $[11, 55]$?

When counting integers in an inclusive range $[a, b]$, the formula is $(b - a) + 1$. If you just subtract $11$ from $55$, you get $44$, which counts the *gaps* between the numbers, not the numbers themselves. By adding $1$, you account for both the starting number $11$ and the ending number $55$.

Question

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

Options:

(A) $\frac{15}{44}$

(B) $\frac{1}{3}$

(C) $\frac{1}{5}$

(D) $\frac{7}{22}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a bridge, looking down at a river. You have two numbers, $x$ and $y$, that must add up to $66$. You want to maximize their product.

If you let $y = 66 - x$, the product becomes:

$$P(x) = x(66 - x)$$

If you were to graph this function, you would see a beautiful, symmetric, downward-opening parabola. The peak of this parabola is the maximum product $M$.

As we know, for a fixed sum, the product is maximized when the numbers are equal. Thus, $x = y = 33$, and our maximum product is:

$$M = 33 	imes 33 = 1089$$

This is the absolute summit of our mathematical landscape.

### The Threshold of Possibility

Now, the problem introduces a constraint: the product must be at least $\frac{5}{9}$ of our maximum $M$. Let us calculate this threshold:

$$\frac{5}{9} 	imes 1089 = 5 	imes 121 = 605$$

We are essentially drawing a horizontal line at $y = 605$ across our parabola. We are interested in the region where the parabola is *above* this line, meaning:

$$x(66 - x) \geq 605$$

This is the region where our product is 'good enough' to be part of our sample space $S$.

### Solving the Inequality

Expanding our condition, we get $66x - x^2 \geq 605$. Rearranging this into a standard quadratic inequality, we have:

$$x^2 - 66x + 605 \leq 0$$

To find the boundaries of our sample space, we solve $x^2 - 66x + 605 = 0$. We need two numbers that multiply to $605$ and add to $66$.

Breaking down $605$, we find $5 	imes 121 = 5 	imes 11 	imes 11 = 55 	imes 11$. Indeed, $55 + 11 = 66$. So, the roots are $11$ and $55$.

The inequality $(x - 11)(x - 55) \leq 0$ tells us that $x$ must lie in the interval $[11, 55]$.

### Counting the Possibilities

Now, we must count the integers in this range. The number of elements in the sample space $S$ is:

$$55 - 11 + 1 = 45$$

These are our $45$ possible values for $x$. Within this set, we look for Event $A$, where $x$ is a multiple of $3$.

The multiples of $3$ in the range $[11, 55]$ are $12, 15, 18, \dots, 54$. This is an arithmetic progression with first term $a = 12$, common difference $d = 3$, and last term $l = 54$.

Using the formula $l = a + (n - 1)d$, we get:

$$54 = 12 + (n - 1)3$$

This simplifies to $42 = (n - 1)3$, so $n - 1 = 14$, and $n = 15$. There are $15$ favorable outcomes.

### The Final Probability

The probability $P(A)$ is the ratio of favorable outcomes to total outcomes:

$$P(A) = \frac{n(A)}{n(S)} = \frac{15}{45}$$

Simplifying this fraction, we arrive at the elegant result:

**$$P(A) = \frac{1}{3}$$**

It is a beautiful conclusion to a journey through the properties of numbers and the geometry of parabolas.

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

Select Answer:

Visualized Solution

Defining the Product Function

Maximizing the Product $M$

Setting the Threshold

Calculating the Threshold Value

Forming the Quadratic Inequality

Finding the Roots

Defining the Valid Range

Counting the Sample Space $n (S)$

Identifying Event $A$

Counting Event Elements $n (A)$

Calculating the Final Probability $P (A)$

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Threshold of Possibility

Solving the Inequality

Counting the Possibilities

The Final Probability

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