Lagrange multipliers - finding maximum/minimum I have solved the question, and obtained the critical points, but don't know how to show its a maximum or minimum of a function. I don't understand other answers because symbols confuse me so much and its still unclear... I would appreciate responses in simple math terms/symbols.
The equation was $f(x, y, z) = xyz$, with constraint $x + y + z = 1$.
After few lines I obtain these C.P:
$(0, 0, 1)$, $(0, 1, 0)$, $(1, 0, 0)$ and $(1/3, 1/3, 1/3)$
When I saw online, the writer just said: 
$f(0, 0, 1) = 0$
$f(0, 1, 0) = 0$
$f(1, 0, 0) = 0 \leftarrow$ Minimums
$f(1/3, 1/3, 1/3) = 1/27 \leftarrow$ Maximum
I'm very confused because I've asked a tutor at my university and he told me to parametrize it and solve. But online it says I have to do Hessian for 3 variables to find whether it is maximum/minimum. (Which we haven't even looked at for 3 variables...!)
How do I know that it's a minimum/maximum? Is it because $1/27$ is bigger than $0$, so it is a maximum?
 A: Using Lagrange Multiplies, we have to solve,
$$\nabla f= \lambda \nabla g$$
$$g(x,y,z)=1$$
This gives the equations,
$$f_x=\lambda g_x \implies yz=\lambda$$
$$f_y=\lambda g_y \implies xz=\lambda$$
$$f_z=\lambda g_z \implies xy=\lambda$$
$$x+y+z=1$$
If you multiply the LHS of each equation by its $x,y,z$ respectively. Then we have,
$$xyz=\lambda(x)$$
$$xyz=\lambda(y)$$
$$xyz=\lambda(z)$$
Notice that if $\lambda = 0$, $xy=zy=xz=0$. This provides the multiple cases
$$f(0,0,1)=0;f(0,1,0)=0;f(1,0,0)=0$$
that the OP suggested. However, without $x,y,z\geq 0$, they do not count as minimums. Because an additional condition is that the region should be closed and bounded set in $\mathbb{R}^3$ otherwise the extreme value theorem won't apply to these values.(This implies that we can't deduce that the largest value is the maximum and the minumum value is the minimum). And $x+y+z=1$ without the additional constraint of $x,y,z>0$ is a plane that is not closed or bounded.
$$\therefore x=y=z$$
Hence,
$$3z=1 \implies z=\dfrac{1}{3}=y=x$$
So, the another of the points is $f(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3})=\dfrac{1}{27}$
I don't know how you got the other points. In this case, it is clear that this value is a maximum. Observe that since there exists a number below $1/27$ (example is $g(1,0,0)$ which gives a value of $f(1,0,0)=0$ which is clearly less than $1/27$) that means it can't be a minimum, so it must be a maximum. I suspect that the solution you got from online has another condition as well that states $x,y,z>0$. This might be the case if they are finding a volume of a rectangular box, $V=xyz$ which can't have negative sides and obviously the minimum volume is $0$.
A: The method of Lagrange multiplier gives $(x,y,z)=(1/3,1/3,1/3)$ as the point of extremum. Since $(x,y,z)=(0,0,1)$ satisfies the constraint and $f(x,y,z)=0 < 1/27=f(1/3,1/3,1/3)$, it's safe to say that $1/27$ is the maximum.
I don't know how you got $(0,0,1),(0,1,0),(1,0,0)$ as critical points-$(1/3,1/3,1/3)$ should be the only point you get using Lagrange multiplier. Note that $f(0,0,1)$ isn't even the minimum since $f(-1,1,1)=-1<0$ and $-1+1+1=1$.
A: For a three variable problem (z,x,y) use the Hessian Matrix: If its determinant is positive, so the critical points give a maximun (relative), and viceversa.
