# Lagrange Multipliers Calculator

## Apply the method of Lagrange multipliers step by step

The calculator will try to find the maxima and minima of the two- or three-variable function, subject to the given constraints, using the method of Lagrange multipliers, with steps shown.

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Find the maximum and minimum values of $f{\left(x,y \right)} = 81 x^{2} + y^{2}$ subject to the constraint $4 x^{2} + y^{2} = 9$.

### Solution

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 $4 x^{2} + y^{2} = 9$ as $4 x^{2} + y^{2} - 9 = 0$.

Form the Lagrangian: $L{\left(x,y,\lambda \right)} = \left(81 x^{2} + y^{2}\right) + \lambda \left(4 x^{2} + y^{2} - 9\right)$.

Find all the first-order partial derivatives:

$\frac{\partial}{\partial x} \left(\left(81 x^{2} + y^{2}\right) + \lambda \left(4 x^{2} + y^{2} - 9\right)\right) = 2 x \left(4 \lambda + 81\right)$ (for steps, see partial derivative calculator).

$\frac{\partial}{\partial y} \left(\left(81 x^{2} + y^{2}\right) + \lambda \left(4 x^{2} + y^{2} - 9\right)\right) = 2 y \left(\lambda + 1\right)$ (for steps, see partial derivative calculator).

$\frac{\partial}{\partial \lambda} \left(\left(81 x^{2} + y^{2}\right) + \lambda \left(4 x^{2} + y^{2} - 9\right)\right) = 4 x^{2} + y^{2} - 9$ (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 $\begin{cases} 2 x \left(4 \lambda + 81\right) = 0 \\ 2 y \left(\lambda + 1\right) = 0 \\ 4 x^{2} + y^{2} - 9 = 0 \end{cases}.$

The system has the following real solutions: $\left(x, y\right) = \left(- \frac{3}{2}, 0\right)$, $\left(x, y\right) = \left(0, -3\right)$, $\left(x, y\right) = \left(0, 3\right)$, $\left(x, y\right) = \left(\frac{3}{2}, 0\right)$.

$f{\left(- \frac{3}{2},0 \right)} = \frac{729}{4}$

$f{\left(0,-3 \right)} = 9$

$f{\left(0,3 \right)} = 9$

$f{\left(\frac{3}{2},0 \right)} = \frac{729}{4}$

Thus, the minimum value is $9$, and the maximum value is $\frac{729}{4}$.

$\frac{729}{4} = 182.25$A at $\left(x, y\right) = \left(- \frac{3}{2}, 0\right) = \left(-1.5, 0\right)$, $\left(x, y\right) = \left(\frac{3}{2}, 0\right) = \left(1.5, 0\right)$A.
$9$A at $\left(x, y\right) = \left(0, -3\right)$, $\left(x, y\right) = \left(0, 3\right)$A.