Question

Suppose \( x, y, z \) are positive real numbers such that \( x + 2y + 3z = 1 \). If \( M \) is the maximum value of \( xyz^2 \), then the value of \( \frac{1}{M} \) is ..................

Hint:

When maximizing a function with a constraint, use Lagrange multipliers to solve the system of equations.

Answer 1

Correct Answer: 1140 - 1160

To find the maximum value of \(xyz^2\) given \(x + 2y + 3z = 1\) with \(x, y, z > 0\), we can use the method of Lagrange multipliers or apply AM-GM inequality. 

We use AM-GM here: 

Given \(x + 2y + 3z = 1\), let's rewrite it as \(x + 2y + 3z = 1\) and note that by the AM-GM inequality, \((x+2y+3z)/6 \geq \sqrt[6]{x \cdot y^2 \cdot z^3}\). This yields \(1/6 \geq \sqrt[6]{x \cdot y^2 \cdot z^3}\). 

Raising both sides to the sixth power gives \(1/6^6 \geq x \cdot y^2 \cdot z^3\) and hence \(1296 \geq \frac{1}{xyz^2}\). 

To achieve equality in AM-GM, the quantities must be equal: let \(x = a, 2y = a, 3z = a\) resulting in \(x = a, y = a/2, z = a/3\). 

Thus, \(a+2(a/2)+3(a/3) = a+a+ a =1\) gives \(a=1/3\), yielding \(x=1/3, y=1/6, z=1/9\). 

We find \(xyz^2 = (1/3)(1/6)((1/9)^2) = 1/1458\), so \(M = 1/1458\) and \(\frac{1}{M} = 1458\) 

Therefore, \(\frac{1}{M} = 1458\).