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in Limit, continuity and differentiability by (41.7k points)

Find the maximum value of x y z for which subject to condition is x2/3 + y2/4 + z2/9 = 1.

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Given x2/3 + y2/4 + z2/9 = 1

Applying, A.M ≥ G.M

  (xyz)  2

Hence, the max value of xyz is 2.

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