Applications of Derivatives
AM-GM / Optimization
Grade 12
Question:
<p>If \(x, y, z\) are positive real numbers, such that \(x + y + z = 1\). If the minimum value of \(\left(1 + \dfrac{1}{x}\right)\left(1 + \dfrac{1}{y}\right)\left(1 + \dfrac{1}{z}\right)\) is K, then find K/10.</p>
Step-by-Step Solution
Key Concept: The expression (1 + 1/x)(1 + 1/y)(1 + 1/z) = (x+1)/x · (y+1)/y · (z+1)/z achieves minimum when x = y = z = 1/3 by AM-GM inequality or Lagrange multipliers, since the constraint x + y + z = 1 forces symmetry at the critical point.
<p><strong>Step 1:</strong> Rewrite the expression: $(1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z}) = \frac{(x+1)(y+1)(z+1)}{xyz}$</p><p><strong>Step 2:</strong> By symmetry and constraint $x + y + z = 1$, the minimum occurs at $x = y = z = \frac{1}{3}$.</p><p><strong>Step 3:</strong> Substitute $x = y = z = \frac{1}{3}$:</p><p>$(1 + 3)(1 + 3)(1 + 3) = 4 \cdot 4 \cdot 4 = 64$</p><p><strong>Step 4:</strong> Therefore $K = 64$, and $\frac{K}{10} = \frac{64}{10} = 6.4$</p><p>∴ Answer: <strong>6.4</strong></p>
Correct Answer: 6.4