Attempt: I know I need to proceed by Lagrange multiplier method, but I am not sure how to proceed after a step we will get the equations as $$x=\lambda x$$$$y=\frac{\lambda}{2} y$$$$z=\frac{\lambda}{3} z$$$$x^2+y^2+z^2=1$$ Now how to solve them? |
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We have: $x^2 = 1 - y^2 - z^2 \implies f(x,y,z) = 1 - y^2 - z^2 + 2y^2 + 3z^2= 1+y^2+2z^2\ge 1$, and this is the minimum of $f$ which is achieved when $y = z = 0, x = \pm 1$. For the max, we have $f_y = 2y, f_z = 4z$, and $f_y = 0 = f_z \implies y = 0 = z$ which is the only critical point in the domain $D = [-1,1]\times [-1,1]= \{(y,z): -1 \le y, z \le 1\}$ of $f$. Thus evaluating the values of $f$ at the boundary points which are $(y,z) = (\pm 1, 0), (0,\pm 1)$, we have the max is $3$ when $x = 0, y = 0, z = \pm 1$. |
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This question is a bit tricky given the case analysis you have to do. Always be mindful about dividing by 0. The first option for $\lambda$ is $\lambda=0$. This gives $x=y=z=0$, which contradicts your constraint equation. So $\lambda\neq 0$. If $x\neq 0$, then $\lambda=1$, giving $y=0,z=0$. With the constraint equation, you get: $(x,y,z)=(1,0,0)$. Notice that $y\neq 0$ and $z\neq 0$ is impossible since both would imply $\lambda=0$, and see reasoning above. There are other cases we've missed, pertaining to other $\lambda$. |
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There are 3 possible values for $\lambda$, and they are $\lambda = 1,2,3$. These values impose 3 cases to our system: Case $\lambda = 1$: if equations 2 and 3 are supposed to be fulfilled $y=z=0$ and the extrema are given by the following equation $x^2=1$ Case $\lambda = 2$: if equations 1 and 3 are supposed to be fulfilled $x=z=0$ and the extrema are given by the following equation $y^2=1$ Case $\lambda = 3$: if equations 1 and 2 are supposed to be fulfilled $x=y=0$ and the extrema are given by the following equation $z^2=1$ I think it is solved that way. Now you have to verify which of them is the absolute maximum and which is the absolute minimum. |
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