MAX,MIN of a 3 variable function : Correctness in writing

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I have to find max,min of $f(x,y,z)=1/(3x^2 +z^2)$constrained on a region $D$



$D =x^2+z^2-y^2-3le 0$ -> $g(x,y,z)=x^2+z^2-y^2-3=0$



  • I used the Lagrange multiplier method: and I got the points

$P_1(pmsqrt3,0,0)$->$f(P_1)=1/9$



$P_2(0,0,pmsqrt3)$->$f(P_2)=1/3$



  • So My question is : (with the fact that $f(x,y,z)$ doesn't contain a $y$ )It is more correct to write the points with $y = 0$ or with $y=y$, like that:

$P_1a(pmsqrty^2+3,0,0)$->$f(P_1a)=1/(3y^2+9)$



$P_2a(0,0,pmsqrty^2+3)$->$f(P_2a)=1/(y^2+3)$




multivariable-calculus lagrange-multiplier




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  • I haven´t checked the calculation, but $y=y$ is the right intuition.
    – callculus
    Jul 26 at 20:23










  • The cost function is not defined for $x=z=0$? It is unbounded above nearby. Furthermore, by appropriate choice of $y$, you can make the cost be any value in $(0,infty)$, so it has no $min$ either. In particular, the $inf$ is zero and the $sup$ is $=infty$. Lagrange is sort of irrelevant here.
    – copper.hat
    Jul 26 at 20:25











  • If in both case you have obtained $y=0$ then we need to set $y=0$.
    – gimusi
    Jul 26 at 20:26










  • my question arises from the fact that the second equation of the Lagrange system is : $0+lambda (-2y)=0$ that means that it is TRUE for $y=0$ or $lambda =0$
    – NPLS
    Jul 26 at 20:27










  • ($x=0$ and $z=0$) is not defined for real numbers
    – NPLS
    Jul 26 at 20:27















up vote
0
down vote

favorite












I have to find max,min of $f(x,y,z)=1/(3x^2 +z^2)$constrained on a region $D$



$D =x^2+z^2-y^2-3le 0$ -> $g(x,y,z)=x^2+z^2-y^2-3=0$



  • I used the Lagrange multiplier method: and I got the points

$P_1(pmsqrt3,0,0)$->$f(P_1)=1/9$



$P_2(0,0,pmsqrt3)$->$f(P_2)=1/3$



  • So My question is : (with the fact that $f(x,y,z)$ doesn't contain a $y$ )It is more correct to write the points with $y = 0$ or with $y=y$, like that:

$P_1a(pmsqrty^2+3,0,0)$->$f(P_1a)=1/(3y^2+9)$



$P_2a(0,0,pmsqrty^2+3)$->$f(P_2a)=1/(y^2+3)$




multivariable-calculus lagrange-multiplier




share|cite|improve this question





















  • I haven´t checked the calculation, but $y=y$ is the right intuition.
    – callculus
    Jul 26 at 20:23










  • The cost function is not defined for $x=z=0$? It is unbounded above nearby. Furthermore, by appropriate choice of $y$, you can make the cost be any value in $(0,infty)$, so it has no $min$ either. In particular, the $inf$ is zero and the $sup$ is $=infty$. Lagrange is sort of irrelevant here.
    – copper.hat
    Jul 26 at 20:25











  • If in both case you have obtained $y=0$ then we need to set $y=0$.
    – gimusi
    Jul 26 at 20:26










  • my question arises from the fact that the second equation of the Lagrange system is : $0+lambda (-2y)=0$ that means that it is TRUE for $y=0$ or $lambda =0$
    – NPLS
    Jul 26 at 20:27










  • ($x=0$ and $z=0$) is not defined for real numbers
    – NPLS
    Jul 26 at 20:27













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have to find max,min of $f(x,y,z)=1/(3x^2 +z^2)$constrained on a region $D$



$D =x^2+z^2-y^2-3le 0$ -> $g(x,y,z)=x^2+z^2-y^2-3=0$



  • I used the Lagrange multiplier method: and I got the points

$P_1(pmsqrt3,0,0)$->$f(P_1)=1/9$



$P_2(0,0,pmsqrt3)$->$f(P_2)=1/3$



  • So My question is : (with the fact that $f(x,y,z)$ doesn't contain a $y$ )It is more correct to write the points with $y = 0$ or with $y=y$, like that:

$P_1a(pmsqrty^2+3,0,0)$->$f(P_1a)=1/(3y^2+9)$



$P_2a(0,0,pmsqrty^2+3)$->$f(P_2a)=1/(y^2+3)$




multivariable-calculus lagrange-multiplier




share|cite|improve this question













I have to find max,min of $f(x,y,z)=1/(3x^2 +z^2)$constrained on a region $D$



$D =x^2+z^2-y^2-3le 0$ -> $g(x,y,z)=x^2+z^2-y^2-3=0$



  • I used the Lagrange multiplier method: and I got the points

$P_1(pmsqrt3,0,0)$->$f(P_1)=1/9$



$P_2(0,0,pmsqrt3)$->$f(P_2)=1/3$



  • So My question is : (with the fact that $f(x,y,z)$ doesn't contain a $y$ )It is more correct to write the points with $y = 0$ or with $y=y$, like that:

$P_1a(pmsqrty^2+3,0,0)$->$f(P_1a)=1/(3y^2+9)$



$P_2a(0,0,pmsqrty^2+3)$->$f(P_2a)=1/(y^2+3)$





multivariable-calculus lagrange-multiplier





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 26 at 20:20









callculus

16.4k31427




16.4k31427









asked Jul 26 at 20:15









NPLS

1819




1819











  • I haven´t checked the calculation, but $y=y$ is the right intuition.
    – callculus
    Jul 26 at 20:23










  • The cost function is not defined for $x=z=0$? It is unbounded above nearby. Furthermore, by appropriate choice of $y$, you can make the cost be any value in $(0,infty)$, so it has no $min$ either. In particular, the $inf$ is zero and the $sup$ is $=infty$. Lagrange is sort of irrelevant here.
    – copper.hat
    Jul 26 at 20:25











  • If in both case you have obtained $y=0$ then we need to set $y=0$.
    – gimusi
    Jul 26 at 20:26










  • my question arises from the fact that the second equation of the Lagrange system is : $0+lambda (-2y)=0$ that means that it is TRUE for $y=0$ or $lambda =0$
    – NPLS
    Jul 26 at 20:27










  • ($x=0$ and $z=0$) is not defined for real numbers
    – NPLS
    Jul 26 at 20:27

















  • I haven´t checked the calculation, but $y=y$ is the right intuition.
    – callculus
    Jul 26 at 20:23










  • The cost function is not defined for $x=z=0$? It is unbounded above nearby. Furthermore, by appropriate choice of $y$, you can make the cost be any value in $(0,infty)$, so it has no $min$ either. In particular, the $inf$ is zero and the $sup$ is $=infty$. Lagrange is sort of irrelevant here.
    – copper.hat
    Jul 26 at 20:25











  • If in both case you have obtained $y=0$ then we need to set $y=0$.
    – gimusi
    Jul 26 at 20:26










  • my question arises from the fact that the second equation of the Lagrange system is : $0+lambda (-2y)=0$ that means that it is TRUE for $y=0$ or $lambda =0$
    – NPLS
    Jul 26 at 20:27










  • ($x=0$ and $z=0$) is not defined for real numbers
    – NPLS
    Jul 26 at 20:27
















I haven´t checked the calculation, but $y=y$ is the right intuition.
– callculus
Jul 26 at 20:23




I haven´t checked the calculation, but $y=y$ is the right intuition.
– callculus
Jul 26 at 20:23












The cost function is not defined for $x=z=0$? It is unbounded above nearby. Furthermore, by appropriate choice of $y$, you can make the cost be any value in $(0,infty)$, so it has no $min$ either. In particular, the $inf$ is zero and the $sup$ is $=infty$. Lagrange is sort of irrelevant here.
– copper.hat
Jul 26 at 20:25





The cost function is not defined for $x=z=0$? It is unbounded above nearby. Furthermore, by appropriate choice of $y$, you can make the cost be any value in $(0,infty)$, so it has no $min$ either. In particular, the $inf$ is zero and the $sup$ is $=infty$. Lagrange is sort of irrelevant here.
– copper.hat
Jul 26 at 20:25













If in both case you have obtained $y=0$ then we need to set $y=0$.
– gimusi
Jul 26 at 20:26




If in both case you have obtained $y=0$ then we need to set $y=0$.
– gimusi
Jul 26 at 20:26












my question arises from the fact that the second equation of the Lagrange system is : $0+lambda (-2y)=0$ that means that it is TRUE for $y=0$ or $lambda =0$
– NPLS
Jul 26 at 20:27




my question arises from the fact that the second equation of the Lagrange system is : $0+lambda (-2y)=0$ that means that it is TRUE for $y=0$ or $lambda =0$
– NPLS
Jul 26 at 20:27












($x=0$ and $z=0$) is not defined for real numbers
– NPLS
Jul 26 at 20:27





($x=0$ and $z=0$) is not defined for real numbers
– NPLS
Jul 26 at 20:27
















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