Question

The maximum value of \(x^2 y z^3\), subject to \(x + y + z = 42\) exist at \((x, y, z) = (\alpha, \beta, \gamma)\), then \(\alpha\beta\gamma =\)

• A. \(6 \times 7^3\)
• B. \(7 \times 6^3\)
• C. \(14 \times 7 \times 21\)
• D. \(12 \times 6 \times 24\)

Hint:

For maximizing \(x^a y^b z^c\) given \(x+y+z=S\), directly write \(x = \frac{a}{a+b+c}S\), \(y = \frac{b}{a+b+c}S\), and \(z = \frac{c}{a+b+c}S\).
This saves valuable time spent writing out the long AM-GM inequalities.

Answer 1

The Correct Option is A

Concept:
• To maximize a product term \(x^a y^b z^c\) subject to a linear sum constraint \(x + y + z = S\), we can utilize the AM-GM inequality.
• We express the sum by breaking the variables into matching parts: \(\underbrace{\frac{x}{a} + \dots + \frac{x}{a}}_{a \text{ times}} + \underbrace{\frac{y}{b} + \dots + \frac{y}{b}}_{b \text{ times}} + \underbrace{\frac{z}{c} + \dots + \frac{z}{c}}_{c \text{ times}}\).
• The maximum value occurs when all these individual sub-terms are completely equal.

Step 1: Set up the equality conditions for maximum value
The target expression is \(x^2 y^1 z^3\), so the exponents are \(a=2, b=1, c=3\).
The total number of parts is \(2 + 1 + 3 = 6\).
For maximum product, the terms must satisfy: \[ \frac{x}{2} = \frac{y}{1} = \frac{z}{3} = k \] Therefore, we get \(x = 2k\), \(y = k\), and \(z = 3k\).

Step 2: Substitute these values into the given constraint to find \(k\)
\[ x + y + z = 42 \] \[ 2k + k + 3k = 42 \] \[ 6k = 42 \implies k = 7 \]

Step 3: Determine the values of \(\alpha, \beta, \gamma\) and calculate \(\alpha\beta\gamma\)
\[ \alpha = x = 2(7) = 14 \] \[ \beta = y = 7 \] \[ \gamma = z = 3(7) = 21 \] Now, compute the product \(\alpha\beta\gamma\): \[ \alpha\beta\gamma = 14 \times 7 \times 21 \] \[ = (2 \times 7) \times 7 \times (3 \times 7) \] \[ = 6 \times 7^3 \]