Max and Min

Calculus Level 4

Find the maximum and minimum values of $x+y+z$ given the following system of equations $\begin{aligned} x+2y+3z&=&1 \\ yz+zx+xy&=&-1 \\ \end{aligned}$

Max :

\frac {3 + \sqrt { 2 } }{ 4 }

Min :

\frac { 3 - \sqrt { 2 } }{ 4 }

Max :

\frac {3 + 3\sqrt { 3 } }{ 2 }

Min :

\frac { 3 - 3\sqrt { 3 } }{ 2 }

Max :

\frac { 3 + 3\sqrt { 3 } }{ 4 }

Min :

\frac { 3 - 3\sqrt { 3 } }{ 4 }

Max :

\frac {1 + \sqrt { 3 } }{ 4 }

Min :

\frac { 1 - \sqrt { 3 } }{ 4 }

1 solution

Nguyen Thanh Long
Dec 31, 2014

$yz+(1-2y-3z)(y+z)+1=0$ $yz+y-2y^2-3yz+z+2yz-3z^2+1=0$ $1-2y^2+z-3z^2+1=0$ $-2y^2+y+z-3z^2+1=0$ $\delta=1+8*(z-3z^2+1)$ $y=\frac{\sqrt{1+8*(z-3z^2+1)}-1)}{-4}$ So We have: $x+y+z=1-y-2z=1-\frac{\sqrt{1+8*(z-3z^2+1)}-1)}{-4}-2z$ With z in the range: [-0.46 .. 0.81] to get the maximum and minimum of x+y+z. $x+y+z_{MAX}=\boxed{\frac{3+3*\sqrt{3}}{4}}$

Hang on, on your second line, the terms of yz should ad up to -4yz and not cancel out. What's the deal with that?

Josh Banister - 6 years, 5 months ago

why not use vectors ? like x+2y+3z=(xi+yj+zk).(i+2j+3k) let xi+yj+zk be some P =modP * root14 cos(angle) 1<= modP root14 x^2 +y^2+z^2 >= 1/14 (x+y+z)^2 -2(xy+yz+xz) >= 1/14 x+y+z >=1/14 - 2 ! what's wrong ?

Vineeth O'Connor - 6 years, 5 months ago

Max and Min

1 solution

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