How do you do? - 2

Algebra Level 5

I = x y z J = x y 2 z K = x y z 3 I = xyz \\ J = xy^2 z \\ K = xyz^3

Find the sum of the maximum values of the expression I I , J J , and K K if positive reals x , y , z x,y,z are constrained by 2 x + 3 y + 4 z = 60 2x + 3y + 4z = 60 .


The answer is 20641.833.

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1 solution

Akul Agrawal
Oct 28, 2015

I 2 x + 3 y + 4 z 3 { ( 2 x ) ( 3 y ) ( 4 z ) } 1 3 20 3 24 x y z I 1000 3 J 2 x + 3 y 2 + 3 y 2 + 4 z 4 { ( 2 x ) ( 3 y 2 ) ( 3 y 2 ) ( 4 z ) } 1 4 15 4 18 x y 2 z J 5625 2 K 2 x + 3 y + 3 4 z 3 5 { ( 2 x ) ( 3 y ) ( 4 z 3 ) 3 } 1 5 12 5 128 9 x y z 3 K 17496 I\\ \quad \quad \quad \frac { 2x+3y+4z }{ 3 } \ge { \{ (2x)(3y)(4z)\} }^{ \frac { 1 }{ 3 } }\\ \Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad { 20 }^{ 3 }\ge 24xyz\\ \Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad \quad I\quad \le \quad \frac { { 1000 } }{ 3 } \\ J\\ \quad \quad \quad \frac { 2x+\frac { 3y }{ 2 } +\frac { 3y }{ 2 } +4z }{ 4 } \ge { \left\{ (2x)(\frac { 3y }{ 2 } )(\frac { 3y }{ 2 } )(4z) \right\} }^{ \frac { 1 }{ 4 } }\\ \Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { 15 }^{ 4 }\ge 18x{ y }^{ 2 }z\\ \Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad J\quad \le \quad \frac { 5625 }{ 2 } \\ K\\ \quad \quad \quad \frac { 2x+3y+3\cdot \frac { 4z }{ 3 } }{ 5 } \ge { \left\{ (2x)(3y){ (\frac { 4z }{ 3 } ) }^{ 3 } \right\} }^{ \frac { 1 }{ 5 } }\\ \Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { 12 }^{ 5 }\ge \frac { 128 }{ 9 } xy{ z }^{ 3 }\\ \Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad K\quad \le \quad 17496

Overrated...? Oh come on... It's stressful...

Manuel Kahayon - 5 years, 6 months ago

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yeah. stressful but easy... for level 5. , i think so.

Akul Agrawal - 5 years, 6 months ago

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Certainly overrated. Fine for lvl 4.

Shreyash Rai - 5 years, 4 months ago

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