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Step-by-Step Detailed Solution
Step 1: Understand the Problem
We have four positive real numbers $a, b, c, d$ with the constraint:
$a + b + c + d = 11.$
We want to maximize the expression $a^5\,b^3\,c^2\,d$. The problem states that the maximum value can be written in the form $3750\,\beta$, and we need to find $\beta$.
Step 2: Recognize the Weighted AM-GM Setup
The exponents in $a^5\,b^3\,c^2\,d$ add up to $5 + 3 + 2 + 1 = 11$, which coincides with the sum $a+b+c+d=11$. This is a strong indication to use the Weighted AM-GM Inequality.
Step 3: Apply the Weighted AM-GM Inequality
The weighted AM-GM inequality for nonnegative real numbers $x_1, x_2, \ldots, x_n$ with positive weights $w_1, w_2, \ldots, w_n$ states:
$
\displaystyle
\frac{w_1\,x_1 \;+\; w_2\,x_2 \;+\; \cdots \;+\; w_n\,x_n}{w_1 + w_2 + \cdots + w_n}
\;\ge\;
\bigl(x_1^{\,w_1}\,x_2^{\,w_2}\,\cdots\,x_n^{\,w_n}\bigr)^{\!\frac{1}{w_1 + w_2 + \cdots + w_n}}.
$
Equality holds if and only if
$
\frac{x_1}{w_1} = \frac{x_2}{w_2} = \cdots = \frac{x_n}{w_n}.
$
Here, identify:
$a$ with weight $5$
$b$ with weight $3$
$c$ with weight $2$
$d$ with weight $1$
Step 4: Express the Equality Condition
From the equality condition, we should have:
$
\frac{a}{5} \;=\; \frac{b}{3} \;=\; \frac{c}{2} \;=\; \frac{d}{1}.
$
Let this common value be $x$. Then:
$
a = 5x,\quad b = 3x,\quad c = 2x,\quad d = x.
$
Step 5: Satisfy the Sum Constraint
Use $a + b + c + d = 11$ to find $x$:
$
5x + 3x + 2x + x = 11 \quad \Longrightarrow \quad 11x = 11 \quad \Longrightarrow \quad x = 1.
$
So, the values of $a, b, c, d$ that maximize the expression are:
$
a = 5,\; b = 3,\; c = 2,\; d = 1.
$
Step 6: Compute the Maximum Value
Substitute these values into $a^5\,b^3\,c^2\,d$:
$
a^5\,b^3\,c^2\,d
= 5^5 \times 3^3 \times 2^2 \times 1.
$
$
5^5 = 3125,\quad 3^3 = 27,\quad 2^2 = 4.
$
$
\therefore\, a^5\,b^3\,c^2\,d = 3125 \times 27 \times 4 = 3125 \times 108 = 337500.
$
Step 7: Relate It to $3750\,\beta$
The problem tells us the maximum value can be written as $3750\,\beta$. Hence:
$
337500 = 3750 \times \beta
\quad \Longrightarrow \quad
\beta = \frac{337500}{3750} = 90.
$
Thus, the value of $\beta$ is 90.
