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Find the minimum and maximum values of $f(x,y,z)=xyz$ given the constraints $x^2+y^2+z^2=1$ and $x+y+z=0$. Does this constraints exist?

i have already found the system $(x-y)(z+2\lambda_1)=0$, $(y-z)(x+2\lambda_1)=0$, $(z-x)(y+2\lambda_1)=0$, $x^2+y^2+z^2-1=0$, $x+y+z=0.$

but i find it difficult to solve the system.

Michael Rozenberg
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Dontor
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4 Answers4

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Note: $(\frac{1}{\sqrt 2} , -\frac{1}{\sqrt 2} , 0)$ and $(\frac{1}{\sqrt 6} , \frac{1}{\sqrt 6} ,- \frac{2}{\sqrt 6})$ satisfy the constraint, and both are orthogonal to one annother.

We could parameterize the constraints.

$x = \frac {1}{\sqrt{2}} \cos \theta + \frac{1}{\sqrt 6} \sin\theta\\ y = -\frac {1}{\sqrt{2}} \cos \theta + \frac{1}{\sqrt 6} \sin\theta\\ z = - \frac{2}{\sqrt 6} \sin\theta$

then we have a problem in one variable.

maximize $(\frac {1}{\sqrt{2}} \cos \theta + \frac{1}{\sqrt 6} \sin\theta)(-\frac {1}{\sqrt{2}} \cos \theta + \frac{1}{\sqrt 6} \sin\theta)(- \frac{2}{\sqrt 6} \sin\theta)$

$(-\frac 12 \cos^2\theta + \frac 16\sin^2\theta)(-\frac{2}{\sqrt 6}\sin\theta)\\ \frac 1{\sqrt 6} \cos^2\theta\sin\theta - \frac 1{3\sqrt 6}\sin^3\theta\\ \frac {\sin 3\theta}{3\sqrt 6}$

This is maximized when $\sin 3\theta = 1,$ and minimized when it equals $-1$

$\pm\frac {1}{3\sqrt 6}$

Calculus is not actually required.

But you wanted to use Lagrange Multipliers.

$F(x,y,z,\lambda,\mu) = xyz - \lambda(x^2+y^2 + z^2 - 1) - \mu(x+y+z)\\ \frac{\partial F}{\partial x} = yz - 2\lambda x -\mu = 0\\ \frac{\partial F}{\partial y} = xz - 2\lambda y -\mu = 0\\ \frac{\partial F}{\partial z} = xy - 2\lambda z -\mu = 0\\ \frac{\partial F}{\partial \lambda} = x^2 + y^2 + z^2 - 1 = 0\\ \frac{\partial F}{\partial \lambda} = x+y+z = 0$

Setting line 1 equal to line 2.

$yz - 2\lambda x = xz - 2\lambda y\\ (x-y) z = 2\lambda (y-x)$

$y = x$ or $-2\lambda = {z}$

first case: $x = y$

$z = -2x\\ x^2 +y^2 + z^2 = x^2 + x^2 + (2x)^2 = 6x^2 = 1\\ (x,y,z) = (\pm \frac{1}{\sqrt{6}}, \pm\frac{1}{\sqrt{6}}, \mp \frac {2}{\sqrt 6})\\ xyz = \pm \frac{1}{3\sqrt 6}$

else $-2\lambda = z$

$yz + zx = \mu\\ xy + z^2 = \mu\\ z^2 -yz - zx +xy = 0\\ (z-x)(z-y) = 0$

And we get a similar result.

user317176
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HINTS

  1. I am assuming the question whether constraints exist really intended to ask if there are any feasible solutions, and you can clearly see that the point $(x,y,z) = (1/\sqrt{2},-1/\sqrt{2},0)$ satisfies them both.
  2. You have $f(x,y,z) = xyz$. Formulate your constraints using functions $g(x,y,z)$ and $h(x,y,z)$. What are these functions?
  3. The Lagrangian is then $$L(x,y,z,\lambda,\mu) = f(x,y,z) + \lambda g(x,y,z) + \mu h(x,y,z)$$ and the extrema can be found by solving the system $\nabla L = 0$ coupled with both constraints in question.

Please show some work, you are welcome to ask for more hints after you show some progress on the problem.

UPDATE

So the Lagrangian is $$ L = xyz + \lambda \left(x^2+y^2+z^2-1\right) + \mu(x+y+z), $$ so the resulting system is $$ \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix} = \nabla L = \begin{pmatrix} yz + 2\lambda x + \mu\\ xz + 2\lambda y + \mu\\ xy + 2\lambda z + \mu \end{pmatrix} $$ Now subtracting the third equation from the second yields $$ 0 = x(z - y) + 2\lambda(y-z) = (z-y)(x - 2\lambda), $$ so either $y=z$ or $x=2\lambda$.

For example, consider $y=z$, then the two constraints imply $x^2+2y^2 = 1$ and $x+2y=0$, so the last equation says $x = -2y$ and the first then means $$ 1 = (-2y)^2+2y^2 \iff y = \pm \sqrt{1/6}. $$ Can you complete this and the other cases?

gt6989b
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By AM-GM we can show that $$-\frac{1}{3\sqrt6}\leq f(x,y,z)\leq\frac{1}{3\sqrt6}.$$ Indeed, let $xy\leq0.$

Thus, $$x^2y^2z^2=4\left(-\frac{xy}{2}\right)^2z^2\leq4\left(\frac{2\left(-\frac{xy}{2}\right)+z^2}{3}\right)^3=$$ $$=4\left(\frac{-xy+(x+y)^2}{3}\right)^3=4\left(\frac{x^2+y^2+z^2}{6}\right)^3=\frac{1}{54}.$$ The equality occurs for $z^2=-\frac{xy}{2},$ which says that we got a maximal and a minimal value of $f$.

Michael Rozenberg
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This is an homogeneous system so after substituting $z = -(x+y)$ we have

$$ {\min(\max) }_{x,y}-x y(x+y) \ \ \text{s. t.}\ \ x^2+y^2+(x+y)^2 = 1 $$

making $y=\lambda x$ and substituting we have the equivalent problem

$$ {\min(\max) }_{\lambda}f(\lambda) = -\frac{\lambda(\lambda+1)}{\left(2(\lambda^2+\lambda+1)\right)^{\frac 32}} $$

Now the stationary condition gives

$$ f'(\lambda) = 0\Rightarrow \sqrt{2} (\lambda -1) (\lambda +2) (2 \lambda +1) = 0 $$

and then

$$ -\frac{1}{3 \sqrt{6}}\le f(\lambda) \le \frac{1}{3 \sqrt{6}} $$

Cesareo
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