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Lagrange multipliers: find maxima and minima of $$$f{\left(x,y \right)} = 81 x^{2} + y^{2}$$$, subject to $$$4 x^{2} + y^{2} = 9$$$

The calculator will try to find the maxima and minima of the multivariable function $$$f{\left(x,y \right)} = 81 x^{2} + y^{2}$$$, subject to the constraint $$$4 x^{2} + y^{2} = 9$$$, 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 $$$\frac{729}{4}$$$.

Answer

Maximum

$$$\frac{729}{4} = 182.25$$$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.

Minimum

$$$9$$$A at $$$\left(x, y\right) = \left(0, -3\right)$$$, $$$\left(x, y\right) = \left(0, 3\right)$$$A.