Optimize $xyz$ where $x+y+z=1$ and $x^2+y^2+z^2=1$? Im trying to optimize
$f(x,y,z)=xyz$ restricted to $g(x,y,z)=x+y+z=1$ and $h(x,y,z)=x^2+y^2+z^2=1$.
$∇f=(yz,xz,xy)$,  $∇g=(1,1,1)$ and $∇h=(2x,2y,2z)$.
I tried using the determinant $det(∇f,∇g,∇h)=yz(2z-2y)-xz(2z-2x)+xy(2y-2x)=0$ which I dont know what to do with and I cant simplify the determinant in a good way with row operations.
I also tried solving $z$ from $g=1$. $z=1-x-y$.
$f(x,y,1-x-y)=xy-x^2y-xy^2$ restricted to $h(x,y)=2x^2+2y^2-2x-2y+2xy+1$ with Lagrange multiplier but I made no progress there either as the partial derivatives got too messy.
 A: Another possible approach is noticing that the intersection of the sphere and the plane produces a circle which can be parametrized by the equation
$$\left(\frac{1-\sqrt{3}\cos(t)+\sin(t)}{3},\frac{1+\sqrt{3}\cos(t)+\sin(t)}{3},\frac{1-2\sin(t)}{3}\right)\qquad t\in[0,2\pi].$$
Now plugging this into the function gives
$$f(t)=-\frac{2}{27}\cdot \left(4 \sin^3(t) - 3 \sin(t) + 1\right)=\frac{2}{27}\cdot \left(\sin(3t)-1\right)\qquad t\in[0,2\pi]$$
which is a function in just one variable that can be simply optimized.
A: $z = 1-x-y\implies f(x,y,z) = xyz = xy(1-x-y)=xy-xy(x+y)=g(x,y)$, subject to: $x^2+y^2+(1-x-y)^2 = 1$, or $2(x^2+y^2)-2x-2y+2xy=0$, or $x^2+y^2-x-y+xy=0$, or $(x+y)^2-xy-(x+y)=0\implies g(x,y) = xy(1-(x+y))=((x+y)^2-(x+y))(1-(x+y))=(x+y)^2-(x+y)^3-(x+y)+(x+y)^2 = -t^3+2t^2-t,t = x+y$. Observe that: $(x+y)^2 - (x+y) = xy \le \dfrac{(x+y)^2}{4}\implies t^2-t \le \dfrac{t^2}{4}\implies 3t^2-4t \le 0 \implies t(3t-4) \le 0\implies 0 \le t  \le \dfrac{4}{3}$. Thus the problem boils down to finding the max and min of $h(t) = -t^3+2t^2-t, 0 \le t \le \dfrac{4}{3}$. You have: $h'(t) = -3t^2+4t-1 = 0\implies (-3t+1)(t-1) = 0\implies t = \dfrac{1}{3}, 1$. Evaluating $h$ at end points and at critical points: $h(0) = 0, h(\frac{4}{3})= -\dfrac{4}{27}, h(1) = 0, h(\frac{1}{3}) = -\dfrac{4}{27}$. This shows that the min value is $-\dfrac{4}{27}$, and the max value is $0$.
A: The Lagrange multiplier method is effective for this problem. First of all observe that the case $x=y=z$ is inadmissible. By symmetry we may assume that $z\neq x$ and $z\neq y.$ We have to solve for
$$\nabla(xyz)=\lambda \nabla(x+y+z)+\mu\,\nabla(x^2+y^2+z^2) $$ i.e.
\begin{eqnarray*}
yz &=& \lambda +2\mu x\\ 
xz &=& \lambda +2\mu y \\
xy &=& \lambda +2\mu z
\end{eqnarray*}
Subtracting the third equation from the first two ones gives
\begin{eqnarray*}y(z-x)=-2\mu(z-x)\\
x(z-y)=-2\mu(z-y)
\end{eqnarray*}
Therefore $x=y.$ We thus get $$2x+z=1,\quad 2x^2+z^2=1$$
This leads to two solutions $x=y=0,$ $z=1$ and $x=y={2\over 3},$ $z=-{1\over 3}.$ Thus
$$M=0,\qquad m={2\over 3}\cdot {2\over 3}\cdot \left (-{1\over 3}\right )=-{4\over 27}$$
A: The intersection locus of the sphere $x^2 + y^2 + z^2 = 1$ and the plane $x + y + z = 1$ is a circle with center at $(\dfrac{1}{3}, \dfrac{1}{3}, \dfrac{1}{3})$ and radius $\sqrt{\dfrac{2}{3}}$ spanned by the orthogonal unit vectors
$u_1 = \dfrac{1}{\sqrt{2}} (1, -1, 0) $
and
$ u_2 = \dfrac{1}{\sqrt{6}} (1, 1, -2 ) $
Therefore, the circle parametrically is given by $q(t) = (x(t), y(t), z(t))$ where
$ x(t) = \dfrac{1}{3} + \dfrac{1}{\sqrt{3}} \cos(t) + \dfrac{1}{3} \sin(t) $
$ y(t) = \dfrac{1}{3} - \dfrac{1}{\sqrt{3}} \cos(t) + \dfrac{1}{3} \sin(t) $
$z(t) = \dfrac{1}{3} - \dfrac{2}{3} \sin(t) $
It follows that
$ x(t) y(t) = \bigg( \dfrac{1}{3} (1 + \sin(t) ) \bigg)^2 - \dfrac{1}{3} \cos^2(t) $
and this simplifies to
$ x(t) y(t) = - \dfrac{2}{9} + \dfrac{2}{9} \sin(t) + \dfrac{4}{9} \sin^2(t) $
Multiplying by $z(t)$ gives
$ f(t) = x(t) y(t) z(t) = \dfrac{2}{27} (-1 + \sin(t) + 2 \sin^2(t) ) ( 1 - 2 \sin(t) ) \\
=  \dfrac{2}{27} (2 \sin(t) - 1)(\sin(t) + 1 )(1 - 2 \sin(t) ) $
Let $g(r) = (2 r - 1)(r + 1)(1 - 2r) = -(r + 1)(4 r^2 - 4 r + 1 ) = - (4 r^3 - 3 r + 1 ) $
$g'(r) = - (12 r^2 - 3 ) = -3 ( 4 r^2 - 1) $
Hence the the critical points are $ r = \pm \dfrac{1}{2} $. And we have  a local minimum at $r = - \dfrac{1}{2} $ where $g(- \dfrac{1}{2}) =  -2 $ and a local maximum at $r = \dfrac{1}{2} $ where $g(\dfrac{1}{2})  = 0 $.  And we also have $g(-1) = 0 $ and $g(1) = -2 $
Based upon all that, we deduce that
The maximum is $0$ and the minimum  is $ \dfrac{-4}{27} $
A: Notice that $$1=1^2=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xy)=1+2(xy+yz+xy)$$
Solving for the products, we learn all elementary symmetric polynomials in $(x,y,z)$: \begin{align*}
x+y+z&=1 \\
xy+xz+yz&=0 \\
xyz&=f
\end{align*} where $f$ is what we want to maximize.  By Vietá's formulae, $x$, $y$, and $z$ are the roots of $$g(t)=t^3-t^2-f=0$$
We must maximize $f$ so that $g$ has $3$ real roots.  To do this, analyze $g$ geometrically.
At local extrema, $$0=g'(t)=3t^2-2t=t(3t-2)$$  There are two extrema: one at $0$ and one at $\frac{2}{3}$.  Since $g(t)$ is increasing at large $|t|$, the former extremum is a maximum; the latter a minimum.
Thus $g$ has three real roots iff $$g(0)\geq0\geq g\left(\frac{2}{3}\right)$$  From the left-hand inequality, $f\leq0$, which is clearly sharp (take $x=y=0$, $z=1$).
