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
 A: If $z$ is the triangle's side on the rotation axis and $h$ is the height of the triangle onto that side, then the center of mass of the triangle is at a distance $$r=\frac 13h$$ from the axis of rotation.
Then the length (cricumference) of the circle drawn by the center of mass during the rotation is $$L=2\pi r = \frac 23\pi h$$
and the triangle's area is $$A=\frac 12 zh$$
so, by the Pappus theorem, the solid's volume is
$$V = A\cdot L = \frac 12\cdot \frac 23 zh\pi h = \frac 13 \pi h^2 z$$
Edit
Now, for any chosen $z$ (which must be less than $a$ for the triangle to exist), we have the maximum $h$ (hence both a maximum $r$ and $L$, and maximum $A$) for $x=y$ (an isosceles triangle). Then
$$x=y=(2a-z)/2=a-\frac z2$$
and
$$h^2 = x^2-(z/2)^2 = a(a - z)$$
Plug that to $V$ to obtain
$$V=\frac 13\pi a(a-z)z$$
which is a quadratic function of $z$.
Its maximum is at the midpoint of its zeros, which are $z=a$ (triangle degenerated to a line segment along the rotation axis) and $z=0$ (triangle degenerated to a line segment perpendicular to the rotation axis):
$$z_{max}=\frac{0+a}2 = a/2$$
and finally
$$V_{max}=\frac 13\pi a(a-z_{max})z_{max} = \frac 13\pi a\cdot\frac a2\cdot\frac a2 = \frac 1{12}\pi a^3$$
A: 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
A: The trick with the formula is that the symbols in this problem do not fit standard symbols used for a cone. 
First note the solid you obtain is a cone only when the triangle is right, with the right angle adjacent to the rotation axis. Otherwise the solid is a union (or a difference) of two cones with a common axis and a common base; in the case of difference the resulting solid is concave. Then the total volume is 
$$V_\text{cone 1}\pm V_\text{cone 2} = \frac 13\pi r_1^2h_1\pm \frac 13\pi r_2^2h_2 =\frac 13\pi r_\text{cone base}^2(h_1\pm h_2).$$ 
And here the trick comes: the radius of the cones' common base was a height of the triangle, whilst the "sum of heights" of obtained cones was a base of the triangle you rotate (i.e. $z$). Hence the confusion:
$$V=\frac 13\pi\ r_\text{cone base}^2\ h_\text{both cones} = \frac 13\pi\ h_\text{triangle}^2\ z.$$
