maximize volume of triangle in given perimeter

Question

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

Answer 1

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 $$V_{max}=\frac 13\pi a(a-z_{max})z_{max} = \frac 13\pi a\cdot\frac a2\cdot\frac a2 = \frac 1{12}\pi a^3$$

Answer 2

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