JEE Main 2016 · 9 Apr · Shift 1 · Mathematics Q68

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MathematicsSingle correct+4 / -1Question 68 of 90

Let x, y, z be positive real numbers such that x + y + z = 12 and $x^3y^4z^5=(0.1)(600)^3$ . Then $x^3+y^3+z^3$ is equal to :

270

258

342

216

Answer and worked solution

Official answer: D

By weighted AM-GM, $x^3y^4z^5$ with $x+y+z=12$ is at most $3^34^45^5=2.16\times10^7$ , reached at x = 3, y = 4, z = 5. The given value $0.1\times600^3=2.16\times10^7$ is this maximum, so $x^3+y^3+z^3=27+64+125=216$ .

Topics:AM-GM-HM inequalities

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JEE Main 2016 · 9 Apr · Shift 1, Mathematics question 68