Volume of parallelepiped problem. Out of all parallelepipeds that have the sum of three dimensions equal to $a$ find the one with greatest volume. Solve this problem using extreme values of multivarible functions.
Sum of all dimensions of paralelopiped is $x+y+z=a$ and it's volume is $V=xyz$ but i don't know how to find the one with largest volume using extreme values of multivariable functions.
 A: use the Lagrange function $$f(x,y,z,\lambda)=xyz+\lambda(x+y+z-a)$$´and compute all partial derivatives with respect to $$x,y,z,\lambda$$
A: The theory behind this is called Lagrange Multipliers.
The usual way to maximize or minimize a function is to find its critical points and analyze them. But here, you don't want to simply find the maximum of $V = xyz$, but you want to find the maximum only among the points that satisfy a certain restriction (given by one equation $g(x,y,z) = a$), in this case $x + y + z = a$.


*

*Find the points $(x,y,z)$ such that the gradient of $V$ is parallel to the gradient of $g$. Those will be the "candidates".

*Manually compare the candidates to see which one is the maximum!

A: The way I learned "Lagrange multipliers" is just a tiny bit different.  To maximize (or minimize) f(x, y, z) with the constraint that g(x, y, z)= C, a constant, find the gradient vectors of both function, $\nabla f$ and $\nabla g$.  One can show that, at a max or min of f, subject to the constraint that g is constant, those two vectors must be parallel.  And that means that there exist some number, $\lambda$, the "Lagrange multiplier", such that $\nabla f= \lambda\nabla g$.  Of course, that gives the same solution as Dr. Sonnhard Graubner's equation.
Here, f(x, y, z)= xyz so $\nabla f= yz\vec{i}+ xz\vec{j}+ xy\vec{k}$ and g(x, y, z)= x+ y+ z= a so $\nabla g= \vec{i}+ \vec{j}+ \vec{k}$.  We have $yz\vec{i}+ xz\vec{j}+ xy\vec{k}= \lambda(\vec{i}+ \vec{j}+ \vec{k})$ so we have the three equations $yz= \lambda$, $xz= \lambda$, and $xy= \lambda$.  Those, together with x+ y+ z= a give four equations to solve for x, y, z, and $\lambda$.  Since a value for $\lambda$ is not necessary for a solution, it is usually simplest just to eliminate $\lambda$ first.  
