Question

Let a, b, c and d be positive real numbers such that a + b + c + d = 11. If the maximum value of \(a^5b ^3c ^2d\) is \(3750\;\beta\), then the value of \(\beta\) is

• A. 110
• B. 90
• C. 55
• D. 108

Answer 1

The Correct Option is B

To maximize the given expression, we use the Arithmetic Mean (A.M.) and Geometric Mean (G.M.) inequality:

A.M. ≥ G.M.

Let us assume the terms as:

\[ \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{b}{3}, \frac{b}{3}, \frac{b}{3}, \frac{c}{2}, \frac{c}{2}, d \]

Now applying A.M. ≥ G.M., we get:

\[ \frac{\frac{a}{5} + \frac{a}{5} + \frac{a}{5} + \frac{a}{5} + \frac{a}{5} + \frac{b}{3} + \frac{b}{3} + \frac{b}{3} + \frac{c}{2} + \frac{c}{2} + d}{11} \geq \sqrt[11]{a^5b^3c^2d} \]

Given \(a + b + c + d = 11\), the left-hand side simplifies to:

\[ \frac{11}{11} \geq \sqrt[11]{a^5b^3c^2d} \]

Therefore:

\[ a^5b^3c^2d \leq 5^5 \cdot 3^3 \cdot 2^2 \cdot 1 \]

Calculating the maximum value:

\[ a^5b^3c^2d \leq 5^5 \cdot 3^3 \cdot 2^2 = 337500 \]

We can express this as:

\[ 337500 = 90 \cdot 3750 \quad \text{where} \, \beta = 90 \]

Explanation

To achieve the maximum value, the numbers must be distributed in the ratios consistent with their powers in \(a^5b^3c^2d\), ensuring the product is maximized. Using A.M. ≥ G.M. is crucial in such optimization problems.