Question 433 of 519MediumMCQ
Let x, y, z be positive real numbers such that x + y + z = 12 and x3y4z5 = (0.1) (600)3. Then x3 + y3 + z3is equal to :
Solutions · 1
Official solution
As we know
AM \[ \ge \] GM
\[ \Rightarrow \] \[{{3\left( {{x \over 3}} \right) + 4\left( {{y \over 4}} \right) + 5\left( {{z \over 5}} \right)} \over {12}}\] \[ \ge \] \[{\left[ {{{\left( {{x \over 3}} \right)}^3}{{\left( {{y \over 4}} \right)}^4}{{\left( {{z \over 5}} \right)}^5}} \right]^{{1 \over {12}}}}\]
\[ \Rightarrow \] 1 \[ \ge \] \[{{{x^3}{y^4}{z^5}} \over {{3^3}{4^4}{5^5}}}\]
\[ \Rightarrow \] x3 y4 z5 \[ \le \] 33 . 44 . 55
\[ \Rightarrow \] x3 y4 z5 \[ \le \] (0.1)(600)3
but given that,
x3 y4 z5 = (0.1) (600)3
\[ \therefore \] AM = GM
\[ \Rightarrow \] All the number are equal.
\[ \therefore \] \[{x \over 3} = {y \over 4} = {z \over 5} = k\]
\[ \Rightarrow \] x = 3k, y = 4k, z = 5k
given that,
x + y + z = 12
\[ \Rightarrow \] 3k + 4k + 5k = 12
\[ \Rightarrow \] 12k = 12
\[ \Rightarrow \] k = 1
\[ \therefore \] x = 3, y = 4, z = 5
So, x3 + y3 + z3
= 33 + 43 + 53
= 216 Answer the question to unlock