Find extrema for $u=x^my^nz^p$ on $x+y+z=a,$ where $a, m, n, p>0.$
Equivalently (as told in class), we investigate $L$ for extrema: $$L=u+\lambda(x+y+z-a).$$
The three partial derivatives are:
$$\begin{align} \\ y^nz^p(mx^{m-1})+\lambda=0\\ x^mz^p(ny^{n-1})+\lambda=0\\ x^my^n(pz^{p-1})+\lambda=0\\ \end{align} $$
From these, $myz=nxz=pxy.$ Expressing every other variable through $x$, then substituting $x, y, z$ in the bond equation (the one with $a$), we find possible points of extrema: $P=(\frac{am}{m+n+p},\frac{an}{m+n+p},\frac{ap}{m+n+p}).$
Now we need to investigate $d^2L$ (a symmetrical matrix made of second-order partial derivatives multiplied by $(dx, dy, dz)^t$, all with values of point P) for its sign.
I've found out (maybe incorrectly) that in terms of sign, $d^2L$ is equivalent to $m+n+p-1$. The book says we have a maximum at $P$. I do not understand how they understood that $m+n+p-1<0$.
Another confusing thing is that, as the book says, to judge by the sign of $d^2L$, we need (in this case) $dx+dy+dz=0$. In $P$, it becomes $a=0$, which breaks the given condition $a>0$.
I'm lost. Need clarification.
Well done! you have correctly done the maths and obtained the extreme point $$(\frac{am}{m+n+p},\frac{an}{m+n+p},\frac{ap}{m+n+p}).$$
For non-negative $x,y,z$, the function $u$ is clearly non-negative. Furthermore, $u$ is positive at the one and only local extreme point $P$ and $u$ is zero at the boundary points (where $xyz=0$).
It therefore requires no complicated analysis to know that $u$ has a maximum at $P$.
If you want a check on your working then please add it clearly to your question but let's emphasise that to prove $P$ gives a maximum requires no work beyond that given above.