Lagrange multipliers: find maxima and minima of $$$f{\left(x,y,z \right)} = x y^{2} z^{3}$$$, subject to $$$x^{2} + y^{2} + z^{2} = 6$$$

Find the maximum and minimum values of $$$f{\left(x,y,z \right)} = x y^{2} z^{3}$$$ subject to the constraint $$$x^{2} + y^{2} + z^{2} = 6$$$.

Attention! This calculator doesn't check the conditions for applying the method of Lagrange multipliers. Use it at your own risk: the answer may be incorrect.

Rewrite the constraint $$$x^{2} + y^{2} + z^{2} = 6$$$ as $$$x^{2} + y^{2} + z^{2} - 6 = 0$$$.

Form the Lagrangian: $$$L{\left(x,y,z,\lambda \right)} = x y^{2} z^{3} + \lambda \left(x^{2} + y^{2} + z^{2} - 6\right)$$$.

Find all the first-order partial derivatives:

$$$\frac{\partial}{\partial x} \left(x y^{2} z^{3} + \lambda \left(x^{2} + y^{2} + z^{2} - 6\right)\right) = 2 \lambda x + y^{2} z^{3}$$$ (for steps, see partial derivative calculator).

$$$\frac{\partial}{\partial y} \left(x y^{2} z^{3} + \lambda \left(x^{2} + y^{2} + z^{2} - 6\right)\right) = 2 y \left(\lambda + x z^{3}\right)$$$ (for steps, see partial derivative calculator).

$$$\frac{\partial}{\partial z} \left(x y^{2} z^{3} + \lambda \left(x^{2} + y^{2} + z^{2} - 6\right)\right) = z \left(2 \lambda + 3 x y^{2} z\right)$$$ (for steps, see partial derivative calculator).

$$$\frac{\partial}{\partial \lambda} \left(x y^{2} z^{3} + \lambda \left(x^{2} + y^{2} + z^{2} - 6\right)\right) = x^{2} + y^{2} + z^{2} - 6$$$ (for steps, see partial derivative calculator).

Next, solve the system $$$\begin{cases} \frac{\partial L}{\partial x} = 0 \\ \frac{\partial L}{\partial y} = 0 \\ \frac{\partial L}{\partial z} = 0 \\ \frac{\partial L}{\partial \lambda} = 0 \end{cases}$$$, or $$$\begin{cases} 2 \lambda x + y^{2} z^{3} = 0 \\ 2 y \left(\lambda + x z^{3}\right) = 0 \\ z \left(2 \lambda + 3 x y^{2} z\right) = 0 \\ x^{2} + y^{2} + z^{2} - 6 = 0 \end{cases}.$$$

The system has the following real solutions: $$$\left(x, y, z\right) = \left(\sqrt{6 - y^{2}}, y, 0\right)$$$, $$$\left(x, y, z\right) = \left(\sqrt{6 - z^{2}}, 0, z\right)$$$, $$$\left(x, y, z\right) = \left(- \sqrt{6 - y^{2}}, y, 0\right)$$$, $$$\left(x, y, z\right) = \left(- \sqrt{6 - z^{2}}, 0, z\right)$$$.

$$$f{\left(\sqrt{6 - y^{2}},y,0 \right)} = 0$$$

$$$f{\left(\sqrt{6 - z^{2}},0,z \right)} = 0$$$

$$$f{\left(- \sqrt{6 - y^{2}},y,0 \right)} = 0$$$

$$$f{\left(- \sqrt{6 - z^{2}},0,z \right)} = 0$$$

As we found only one value, you will have to check yet whether it is the maximum or the minimum. To do this, take another point satisfying the constraint(s) and find the value of the function at it. If the value at this new point is less than the value at the original point, then the original point is the maximum. On the contrary, if the value at the new point is greater, then the original point is the minimum.

Can't find the maximum and minimum.