Question:
Find the greatest value of \(xyz\) for positive values of \(x,y,z\) subject to the condition \(xy + yz + zx = 12\)
Find the greatest value of \(xyz\) for positive values of \(x,y,z\) subject to the condition \(xy + yz + zx = 12\)
Show Hint
Use AM-GM inequality on the three product terms.
Updated On: Apr 23, 2026
- 64
- 8
- 16
- 32
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The Correct Option is B
Solution and Explanation
Step 1: Formula / Definition}
\[ \text{AM-GM: } \frac{xy + yz + zx}{3} \geq \sqrt[3]{(xy)(yz)(zx)} \]
Step 2: Calculation / Simplification}
\(\frac{12}{3} \geq \sqrt[3]{x^2 y^2 z^2} = (xyz)^{2/3}\)
\(4 \geq (xyz)^{2/3} \Rightarrow (xyz)^{2/3} \leq 4\)
\(xyz \leq 4^{3/2} = 8\)
Equality when \(xy = yz = zx = 4 \Rightarrow x = y = z = 2\)
Step 3: Final Answer
\[ 8 \]
\[ \text{AM-GM: } \frac{xy + yz + zx}{3} \geq \sqrt[3]{(xy)(yz)(zx)} \]
Step 2: Calculation / Simplification}
\(\frac{12}{3} \geq \sqrt[3]{x^2 y^2 z^2} = (xyz)^{2/3}\)
\(4 \geq (xyz)^{2/3} \Rightarrow (xyz)^{2/3} \leq 4\)
\(xyz \leq 4^{3/2} = 8\)
Equality when \(xy = yz = zx = 4 \Rightarrow x = y = z = 2\)
Step 3: Final Answer
\[ 8 \]
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