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.