Lagrange multipliers (min and max with 2 constraints)

Question

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.

Answer 1

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.

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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?

Answer 2

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 $f$.

Answer 3

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.

Answer 4

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

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