<s>[INST] maximize volume of triangle in given perimeter question: Find the triangle with perimeter 2a given that, when we rotate it around one of its sides, the solid obtained have the maximum volume.
suppose we have a triangle of sides x,y,z such that $x+y+z=2a$ and $h$ is the height,  we rotate it around the side with size z, we want to get the solid with maximum volume. $Area = \sqrt{a(a-y)(a-y)(a-z)}=\frac{1}{2} zh$
but i was confuse why the volume is $(\frac{1}{3}\pi h^2z)$
 if it is cone why not  $(\frac{1}{3}\pi z^2h)$??
and is it better to use Lagrange multipliers or substitution method? thankyou
 [/INST] From $Area=\frac12 \sqrt{a (a-x) (a-y) (a-z)}$
we get $h=\dfrac{2 \sqrt{a (a-x) (a-y) (a-z)}}{z}$
Plug in the volume $V=\frac{1}{3} \pi  h^2 z$ and get
$V=\dfrac{4 \pi  a (a-x) (a-y) (a-z)}{3 z}$
with the constraint $x+y+z-2a=0$
We use the Lagrangian multiplier and consider
$f(x,y,z,k)=\dfrac{4 \pi  a (a-x) (a-y) (a-z)}{3 z}+k (-2 a+x+y+z)$
Derivative wrt $x,y,z$ must be zero
$$
\left\{ {\begin{array}{*{20}{l}}
  { k-\dfrac{4 \pi  a (a-y) (a-z)}{3 z}=0} \\ 
  {k-\dfrac{4 \pi  a (a-x) (a-z)}{3 z}=0 } \\ 
  {k-\dfrac{4 \pi  a (a-x) (a-y) (a-2 z)}{3 z^2}=0 } 
\end{array}} \right.
$$
we get the solution
$$x=\frac{4 \pi  a^2-2 \sqrt{3 \pi } a \sqrt{k}+3 k}{4 \pi  a},\;y= \frac{4 \pi  a^2-2 \sqrt{3 \pi } a \sqrt{k}+3 k}{4 \pi  a},\;z= \frac{2 \pi  a-\sqrt{3 \pi } \sqrt{k}}{2 \pi }$$
Now we must consider the constraint so we substitute in the equation
$x+y+z=2a$ and we get
$$\frac{4 \pi  a^2+2 \sqrt{3 \pi } a \sqrt{k}+3 k}{2 \pi  a}+\frac{2 \pi  a+\sqrt{3 \pi } \sqrt{k}}{2 \pi }-2 a=0$$
there are two solutions $\quad k=\dfrac{\pi  a^2}{3},\;k= \dfrac{4 \pi  a^2}{3}$
but only the first leads to the result which is
$x=\dfrac{3 a}{4},\;y=\dfrac{3 a}{4},\;z=\dfrac{a}{2}$
and a volume $V=\dfrac{\pi  }{12}\,a^3$
I would have bet that the maximum was the equilateral triangle, but solid generated has a volume smaller, namely $V'=\dfrac{2 \pi }{27}\,a^3$
hope this helps
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