Lagrange Multipliers Calculator

Find the maximum and minimum values of $$$f{\left(x,y \right)} = 3 x + 4 y$$$ subject to the constraint $$$x^{2} + y^{2} = 25$$$.

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 $$$x^{2} + y^{2} = 25$$$ as $$$x^{2} + y^{2} - 25 = 0$$$.

Form the Lagrangian: $$$L{\left(x,y,\lambda \right)} = \left(3 x + 4 y\right) + \lambda \left(x^{2} + y^{2} - 25\right)$$$.

Find all the first-order partial derivatives:

$$$\frac{\partial}{\partial x} \left(\left(3 x + 4 y\right) + \lambda \left(x^{2} + y^{2} - 25\right)\right) = 2 \lambda x + 3$$$ (for steps, see partial derivative calculator).

$$$\frac{\partial}{\partial y} \left(\left(3 x + 4 y\right) + \lambda \left(x^{2} + y^{2} - 25\right)\right) = 2 \lambda y + 4$$$ (for steps, see partial derivative calculator).

$$$\frac{\partial}{\partial \lambda} \left(\left(3 x + 4 y\right) + \lambda \left(x^{2} + y^{2} - 25\right)\right) = x^{2} + y^{2} - 25$$$ (for steps, see partial derivative calculator).

Next, solve the system $$$\begin{cases} \frac{\partial L}{\partial x} = 0 \\ \frac{\partial L}{\partial y} = 0 \\ \frac{\partial L}{\partial \lambda} = 0 \end{cases}$$$, or $$$\begin{cases} 2 \lambda x + 3 = 0 \\ 2 \lambda y + 4 = 0 \\ x^{2} + y^{2} - 25 = 0 \end{cases}$$$.

The system has the following real solutions: $$$\left(x, y\right) = \left(-3, -4\right)$$$, $$$\left(x, y\right) = \left(3, 4\right)$$$.

$$$f{\left(-3,-4 \right)} = -25$$$

$$$f{\left(3,4 \right)} = 25$$$

Thus, the minimum value is $$$-25$$$, and the maximum value is $$$25$$$.

Maximum

$$$25$$$A at $$$\left(x, y\right) = \left(3, 4\right)$$$A.

Minimum

$$$-25$$$A at $$$\left(x, y\right) = \left(-3, -4\right)$$$A.