Find any local max or min of $x^2+y^2+z^2$ s.t $x+y+z=1$ and $3x+y+z=5$












3












$begingroup$



Find any local max or min of
begin{align}
f(x,y,z)=x^2+y^2+z^2 && (1)
end{align}
such that
begin{align}
x+y+z=1 && (2)\
3x+y+z=5 && (3)
end{align}




My attempt. Let
$L(x,y,z,lambda_1, lambda_2)= f(x,y,z)+lambda_2 (x+y+z-1) + lambda_1 (3x+y+z-5)$



$L_x=2x+ 3 lambda_1 + lambda_2 =0$



$L_y=2y+ lambda_1 + lambda_2=0$



$L_z=2z+lambda_1 + lambda_2=0$



Solve for $x,y,z$ we get:



$x=frac{-3 lambda_1 - lambda_2}{2}$



$z=y=frac{-lambda_1 - lambda_2}{2}$



with the use of $(2)$ and $(3)$ $implies$



$x=2$



$y=z= frac{-1}{2}$



so the stationary point is $(x,y,z)=(2, frac{-1}{2},frac{-1}{2})$



The Hessian of $L$ gives a postive definite matrix for all $(x,y,z)$ thus



$(x,y,z)=(2, frac{-1}{2},frac{-1}{2})$ is the only local minimizer of $f$ and there is no maximizors of $f$.



Is this correct?






calculus multivariable-calculus optimization nonlinear-optimization lagrange-multiplier







share|cite|improve this question











$endgroup$












  • $begingroup$
    I do not see any errors. I would say that is correct.
    $endgroup$
    – idriskameni
    Jan 17 at 13:08










  • $begingroup$
    It is fine for me.
    $endgroup$
    – Alex Silva
    Jan 17 at 13:10
















3












$begingroup$



Find any local max or min of
begin{align}
f(x,y,z)=x^2+y^2+z^2 && (1)
end{align}
such that
begin{align}
x+y+z=1 && (2)\
3x+y+z=5 && (3)
end{align}




My attempt. Let
$L(x,y,z,lambda_1, lambda_2)= f(x,y,z)+lambda_2 (x+y+z-1) + lambda_1 (3x+y+z-5)$



$L_x=2x+ 3 lambda_1 + lambda_2 =0$



$L_y=2y+ lambda_1 + lambda_2=0$



$L_z=2z+lambda_1 + lambda_2=0$



Solve for $x,y,z$ we get:



$x=frac{-3 lambda_1 - lambda_2}{2}$



$z=y=frac{-lambda_1 - lambda_2}{2}$



with the use of $(2)$ and $(3)$ $implies$



$x=2$



$y=z= frac{-1}{2}$



so the stationary point is $(x,y,z)=(2, frac{-1}{2},frac{-1}{2})$



The Hessian of $L$ gives a postive definite matrix for all $(x,y,z)$ thus



$(x,y,z)=(2, frac{-1}{2},frac{-1}{2})$ is the only local minimizer of $f$ and there is no maximizors of $f$.



Is this correct?






calculus multivariable-calculus optimization nonlinear-optimization lagrange-multiplier







share|cite|improve this question











$endgroup$












  • $begingroup$
    I do not see any errors. I would say that is correct.
    $endgroup$
    – idriskameni
    Jan 17 at 13:08










  • $begingroup$
    It is fine for me.
    $endgroup$
    – Alex Silva
    Jan 17 at 13:10














3












3








3





$begingroup$



Find any local max or min of
begin{align}
f(x,y,z)=x^2+y^2+z^2 && (1)
end{align}
such that
begin{align}
x+y+z=1 && (2)\
3x+y+z=5 && (3)
end{align}




My attempt. Let
$L(x,y,z,lambda_1, lambda_2)= f(x,y,z)+lambda_2 (x+y+z-1) + lambda_1 (3x+y+z-5)$



$L_x=2x+ 3 lambda_1 + lambda_2 =0$



$L_y=2y+ lambda_1 + lambda_2=0$



$L_z=2z+lambda_1 + lambda_2=0$



Solve for $x,y,z$ we get:



$x=frac{-3 lambda_1 - lambda_2}{2}$



$z=y=frac{-lambda_1 - lambda_2}{2}$



with the use of $(2)$ and $(3)$ $implies$



$x=2$



$y=z= frac{-1}{2}$



so the stationary point is $(x,y,z)=(2, frac{-1}{2},frac{-1}{2})$



The Hessian of $L$ gives a postive definite matrix for all $(x,y,z)$ thus



$(x,y,z)=(2, frac{-1}{2},frac{-1}{2})$ is the only local minimizer of $f$ and there is no maximizors of $f$.



Is this correct?






calculus multivariable-calculus optimization nonlinear-optimization lagrange-multiplier







share|cite|improve this question











$endgroup$





Find any local max or min of
begin{align}
f(x,y,z)=x^2+y^2+z^2 && (1)
end{align}
such that
begin{align}
x+y+z=1 && (2)\
3x+y+z=5 && (3)
end{align}




My attempt. Let
$L(x,y,z,lambda_1, lambda_2)= f(x,y,z)+lambda_2 (x+y+z-1) + lambda_1 (3x+y+z-5)$



$L_x=2x+ 3 lambda_1 + lambda_2 =0$



$L_y=2y+ lambda_1 + lambda_2=0$



$L_z=2z+lambda_1 + lambda_2=0$



Solve for $x,y,z$ we get:



$x=frac{-3 lambda_1 - lambda_2}{2}$



$z=y=frac{-lambda_1 - lambda_2}{2}$



with the use of $(2)$ and $(3)$ $implies$



$x=2$



$y=z= frac{-1}{2}$



so the stationary point is $(x,y,z)=(2, frac{-1}{2},frac{-1}{2})$



The Hessian of $L$ gives a postive definite matrix for all $(x,y,z)$ thus



$(x,y,z)=(2, frac{-1}{2},frac{-1}{2})$ is the only local minimizer of $f$ and there is no maximizors of $f$.



Is this correct?






calculus multivariable-calculus optimization nonlinear-optimization lagrange-multiplier




calculus multivariable-calculus optimization nonlinear-optimization lagrange-multiplier






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 13:29









Robert Z

98.6k1068139




98.6k1068139










asked Jan 17 at 13:00









Dreamer123Dreamer123

32729




32729












  • $begingroup$
    I do not see any errors. I would say that is correct.
    $endgroup$
    – idriskameni
    Jan 17 at 13:08










  • $begingroup$
    It is fine for me.
    $endgroup$
    – Alex Silva
    Jan 17 at 13:10


















  • $begingroup$
    I do not see any errors. I would say that is correct.
    $endgroup$
    – idriskameni
    Jan 17 at 13:08










  • $begingroup$
    It is fine for me.
    $endgroup$
    – Alex Silva
    Jan 17 at 13:10
















$begingroup$
I do not see any errors. I would say that is correct.
$endgroup$
– idriskameni
Jan 17 at 13:08




$begingroup$
I do not see any errors. I would say that is correct.
$endgroup$
– idriskameni
Jan 17 at 13:08












$begingroup$
It is fine for me.
$endgroup$
– Alex Silva
Jan 17 at 13:10




$begingroup$
It is fine for me.
$endgroup$
– Alex Silva
Jan 17 at 13:10










2 Answers
2






active

oldest

votes


















1












$begingroup$

Your result is correct. This is an alternative approach.



Here we have two non-parallel planes which intersect along the line $(x,y,z)=(2,t,-1-t)$ where $tin mathbb{R}$.
Hence we reduce the problem to a $1$-variable case,
$$f(x,y,z)=4+t^2+(-1-t)^2=2t^2+2t+5$$
The above quadratic polynomial has just one local/global minimum at $t=-1/2$ that is at $(2,-1/2,-1/2)$. There is no local/global maximum.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great answer. Thank you!
    $endgroup$
    – Dreamer123
    Jan 17 at 13:33



















1












$begingroup$

An option:



1) $x+y+z=1$; and



2) $3x+y+z=5$;



$2$ planes , their intersection is a straight line.



Subtract: 2)-1):



$2x=4$; $x=2$ ;and



$y+z=-1$;



$d^2=x^2+y^2+z^2$ .



Minimal distance of line from origin:



$d^2= 4 +y^2+z^2.$



2D problem:



Minimize $y^2+z^2$ with constraint $y+z=-1$.



$d_2^2= $



$[-(1+z)]^2+z^2=2z^2+2z+1=$



$2(z^2+z)+1= $



$2[(z+1/2)^2]-1/2+1ge 1/2$.



Equality at $z=-1/2$;



Finally:



Minimum at :



$x=2$ ; $y=-1/2$; $z=-1/2$;



$d^2_{min}= 4+1/2=9/2;$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You mean $x=2$?
    $endgroup$
    – Alex Silva
    Jan 17 at 13:48










  • $begingroup$
    Alex.Thanks, this was fast:)
    $endgroup$
    – Peter Szilas
    Jan 17 at 13:50











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Your result is correct. This is an alternative approach.



Here we have two non-parallel planes which intersect along the line $(x,y,z)=(2,t,-1-t)$ where $tin mathbb{R}$.
Hence we reduce the problem to a $1$-variable case,
$$f(x,y,z)=4+t^2+(-1-t)^2=2t^2+2t+5$$
The above quadratic polynomial has just one local/global minimum at $t=-1/2$ that is at $(2,-1/2,-1/2)$. There is no local/global maximum.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great answer. Thank you!
    $endgroup$
    – Dreamer123
    Jan 17 at 13:33
















1












$begingroup$

Your result is correct. This is an alternative approach.



Here we have two non-parallel planes which intersect along the line $(x,y,z)=(2,t,-1-t)$ where $tin mathbb{R}$.
Hence we reduce the problem to a $1$-variable case,
$$f(x,y,z)=4+t^2+(-1-t)^2=2t^2+2t+5$$
The above quadratic polynomial has just one local/global minimum at $t=-1/2$ that is at $(2,-1/2,-1/2)$. There is no local/global maximum.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great answer. Thank you!
    $endgroup$
    – Dreamer123
    Jan 17 at 13:33














1












1








1





$begingroup$

Your result is correct. This is an alternative approach.



Here we have two non-parallel planes which intersect along the line $(x,y,z)=(2,t,-1-t)$ where $tin mathbb{R}$.
Hence we reduce the problem to a $1$-variable case,
$$f(x,y,z)=4+t^2+(-1-t)^2=2t^2+2t+5$$
The above quadratic polynomial has just one local/global minimum at $t=-1/2$ that is at $(2,-1/2,-1/2)$. There is no local/global maximum.






share|cite|improve this answer











$endgroup$



Your result is correct. This is an alternative approach.



Here we have two non-parallel planes which intersect along the line $(x,y,z)=(2,t,-1-t)$ where $tin mathbb{R}$.
Hence we reduce the problem to a $1$-variable case,
$$f(x,y,z)=4+t^2+(-1-t)^2=2t^2+2t+5$$
The above quadratic polynomial has just one local/global minimum at $t=-1/2$ that is at $(2,-1/2,-1/2)$. There is no local/global maximum.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 17 at 13:28

























answered Jan 17 at 13:11









Robert ZRobert Z

98.6k1068139




98.6k1068139












  • $begingroup$
    Great answer. Thank you!
    $endgroup$
    – Dreamer123
    Jan 17 at 13:33


















  • $begingroup$
    Great answer. Thank you!
    $endgroup$
    – Dreamer123
    Jan 17 at 13:33
















$begingroup$
Great answer. Thank you!
$endgroup$
– Dreamer123
Jan 17 at 13:33




$begingroup$
Great answer. Thank you!
$endgroup$
– Dreamer123
Jan 17 at 13:33











1












$begingroup$

An option:



1) $x+y+z=1$; and



2) $3x+y+z=5$;



$2$ planes , their intersection is a straight line.



Subtract: 2)-1):



$2x=4$; $x=2$ ;and



$y+z=-1$;



$d^2=x^2+y^2+z^2$ .



Minimal distance of line from origin:



$d^2= 4 +y^2+z^2.$



2D problem:



Minimize $y^2+z^2$ with constraint $y+z=-1$.



$d_2^2= $



$[-(1+z)]^2+z^2=2z^2+2z+1=$



$2(z^2+z)+1= $



$2[(z+1/2)^2]-1/2+1ge 1/2$.



Equality at $z=-1/2$;



Finally:



Minimum at :



$x=2$ ; $y=-1/2$; $z=-1/2$;



$d^2_{min}= 4+1/2=9/2;$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You mean $x=2$?
    $endgroup$
    – Alex Silva
    Jan 17 at 13:48










  • $begingroup$
    Alex.Thanks, this was fast:)
    $endgroup$
    – Peter Szilas
    Jan 17 at 13:50
















1












$begingroup$

An option:



1) $x+y+z=1$; and



2) $3x+y+z=5$;



$2$ planes , their intersection is a straight line.



Subtract: 2)-1):



$2x=4$; $x=2$ ;and



$y+z=-1$;



$d^2=x^2+y^2+z^2$ .



Minimal distance of line from origin:



$d^2= 4 +y^2+z^2.$



2D problem:



Minimize $y^2+z^2$ with constraint $y+z=-1$.



$d_2^2= $



$[-(1+z)]^2+z^2=2z^2+2z+1=$



$2(z^2+z)+1= $



$2[(z+1/2)^2]-1/2+1ge 1/2$.



Equality at $z=-1/2$;



Finally:



Minimum at :



$x=2$ ; $y=-1/2$; $z=-1/2$;



$d^2_{min}= 4+1/2=9/2;$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You mean $x=2$?
    $endgroup$
    – Alex Silva
    Jan 17 at 13:48










  • $begingroup$
    Alex.Thanks, this was fast:)
    $endgroup$
    – Peter Szilas
    Jan 17 at 13:50














1












1








1





$begingroup$

An option:



1) $x+y+z=1$; and



2) $3x+y+z=5$;



$2$ planes , their intersection is a straight line.



Subtract: 2)-1):



$2x=4$; $x=2$ ;and



$y+z=-1$;



$d^2=x^2+y^2+z^2$ .



Minimal distance of line from origin:



$d^2= 4 +y^2+z^2.$



2D problem:



Minimize $y^2+z^2$ with constraint $y+z=-1$.



$d_2^2= $



$[-(1+z)]^2+z^2=2z^2+2z+1=$



$2(z^2+z)+1= $



$2[(z+1/2)^2]-1/2+1ge 1/2$.



Equality at $z=-1/2$;



Finally:



Minimum at :



$x=2$ ; $y=-1/2$; $z=-1/2$;



$d^2_{min}= 4+1/2=9/2;$






share|cite|improve this answer











$endgroup$



An option:



1) $x+y+z=1$; and



2) $3x+y+z=5$;



$2$ planes , their intersection is a straight line.



Subtract: 2)-1):



$2x=4$; $x=2$ ;and



$y+z=-1$;



$d^2=x^2+y^2+z^2$ .



Minimal distance of line from origin:



$d^2= 4 +y^2+z^2.$



2D problem:



Minimize $y^2+z^2$ with constraint $y+z=-1$.



$d_2^2= $



$[-(1+z)]^2+z^2=2z^2+2z+1=$



$2(z^2+z)+1= $



$2[(z+1/2)^2]-1/2+1ge 1/2$.



Equality at $z=-1/2$;



Finally:



Minimum at :



$x=2$ ; $y=-1/2$; $z=-1/2$;



$d^2_{min}= 4+1/2=9/2;$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 17 at 13:49

























answered Jan 17 at 13:45









Peter SzilasPeter Szilas

11.4k2822




11.4k2822












  • $begingroup$
    You mean $x=2$?
    $endgroup$
    – Alex Silva
    Jan 17 at 13:48










  • $begingroup$
    Alex.Thanks, this was fast:)
    $endgroup$
    – Peter Szilas
    Jan 17 at 13:50


















  • $begingroup$
    You mean $x=2$?
    $endgroup$
    – Alex Silva
    Jan 17 at 13:48










  • $begingroup$
    Alex.Thanks, this was fast:)
    $endgroup$
    – Peter Szilas
    Jan 17 at 13:50
















$begingroup$
You mean $x=2$?
$endgroup$
– Alex Silva
Jan 17 at 13:48




$begingroup$
You mean $x=2$?
$endgroup$
– Alex Silva
Jan 17 at 13:48












$begingroup$
Alex.Thanks, this was fast:)
$endgroup$
– Peter Szilas
Jan 17 at 13:50




$begingroup$
Alex.Thanks, this was fast:)
$endgroup$
– Peter Szilas
Jan 17 at 13:50


















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