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Monday, May 19, 2008

Q) Let x, y and z be three positive numbers such that x + y + z = 12. Then the maximum value of x2y3z is ?

A) x+y+z = 12

x/2 + x/2 + y/3 + y/3 + y/3 + z = 12

Use AM > GM inequality:

(x + y + z) / 6 > [ (x^2 / 4)(y^3 / 27)(z) ]^1/6
2 > [ (x^2 / 4)(y^3 / 27)(z) ]^1/6
2^6 > (x^2 / 4)(y^3 / 27)(z)
2^6 * 4*27 > x^2.y^3.z

So x2y3z < 6912

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