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Find max(x + y + z)*min(x + y + z)

Albert

Well-known member
Jan 25, 2013
1,225
$x,y,z$ are non-negative real numbers ,given:

$4^\sqrt{5x+9y+4z}-68\times2^\sqrt{5x+9y+4z}+256=0$

please find:

$\max(x+y+z)\times \min(x+y+z)$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Re: find max(x+y+z)*min(x+y+z)

Let $\lambda = 2^\sqrt{5x+9y+4z}$. Then the equation becomes $\lambda^2 - 68 \lambda + 256 = 0$, with solutions $\lambda = 4 = 2^2$ and $\lambda = 64 = 2^6.$ Thus $2^\sqrt{5x+9y+4z} = 2^2$ or $2^6$, and so $5x+9y+4z = 4$ or $36$. When $5x+9y+4z = 4$ we get $(x+y+z)_{\min} = 4/9$ by taking $(x,y,z) = (0,4/9,0)$. When $5x+9y+4z = 36$ we get $(x+y+z)_{\max} = 9$ by taking $(x,y,z) = (0,0,9)$. Therefore $(x+y+z)_{\min}\times (x+y+z)_{\max} = 4$.
 

Albert

Well-known member
Jan 25, 2013
1,225
Re: find max(x+y+z)*min(x+y+z)

Let $\lambda = 2^\sqrt{5x+9y+4z}$. Then the equation becomes $\lambda^2 - 68 \lambda + 256 = 0$, with solutions $\lambda = 4 = 2^2$ and $\lambda = 64 = 2^6.$ Thus $2^\sqrt{5x+9y+4z} = 2^2$ or $2^6$, and so $5x+9y+4z = 4$ or $36$. When $5x+9y+4z = 4$ we get $(x+y+z)_{\min} = 4/9$ by taking $(x,y,z) = (0,4/9,0)$. When $5x+9y+4z = 36$ we get $(x+y+z)_{\max} = 9$ by taking $(x,y,z) = (0,0,9)$. Therefore $(x+y+z)_{\min}\times (x+y+z)_{\max} = 4$.

from Opalg's mention :
$5x+9y+4z = 4$ or $36$.
for:$4x+4y+4z\leq 5x+4y+9z\leq 9x+9y+9z$
if $5x+4y+9z=4$
then :$4x+4y+4z\leq 4 \leq 9x+9y+9z$
$\therefore x+y+z\geq \dfrac {4}{9}$
if $5x+4y+9z=36$
then :$4x+4y+4z\leq 36 \leq 9x+9y+9z$
$\therefore x+y+z\leq 9$
and we get :$max(x+y+z)\times min(x+y+z)=4$