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Lagrange multipliers: find maxima and minima of $$$f{\left(x,y \right)} = 4 x + y$$$, subject to $$$20 = \frac{5 \sqrt{x} \sqrt{y}}{4}$$$
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Your Input
Find the maximum and minimum values of $$$f{\left(x,y \right)} = 4 x + y$$$ subject to the constraint $$$20 = \frac{5 \sqrt{x} \sqrt{y}}{4}$$$.
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 $$$20 = \frac{5 \sqrt{x} \sqrt{y}}{4}$$$ as $$$- \frac{5 \sqrt{x} \sqrt{y}}{4} + 20 = 0$$$.
Form the Lagrangian: $$$L{\left(x,y,\lambda \right)} = \left(4 x + y\right) + \lambda \left(- \frac{5 \sqrt{x} \sqrt{y}}{4} + 20\right)$$$.
Find all the first-order partial derivatives:
$$$\frac{\partial}{\partial x} \left(\left(4 x + y\right) + \lambda \left(- \frac{5 \sqrt{x} \sqrt{y}}{4} + 20\right)\right) = - \frac{5 \lambda \sqrt{y}}{8 \sqrt{x}} + 4$$$ (for steps, see partial derivative calculator).
$$$\frac{\partial}{\partial y} \left(\left(4 x + y\right) + \lambda \left(- \frac{5 \sqrt{x} \sqrt{y}}{4} + 20\right)\right) = - \frac{5 \lambda \sqrt{x}}{8 \sqrt{y}} + 1$$$ (for steps, see partial derivative calculator).
$$$\frac{\partial}{\partial \lambda} \left(\left(4 x + y\right) + \lambda \left(- \frac{5 \sqrt{x} \sqrt{y}}{4} + 20\right)\right) = - \frac{5 \sqrt{x} \sqrt{y}}{4} + 20$$$ (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 $$$\begin{cases} - \frac{5 \lambda \sqrt{y}}{8 \sqrt{x}} + 4 = 0 \\ - \frac{5 \lambda \sqrt{x}}{8 \sqrt{y}} + 1 = 0 \\ - \frac{5 \sqrt{x} \sqrt{y}}{4} + 20 = 0 \end{cases}.$$$
The system has the following real solution: $$$\left(x, y\right) = \left(8, 32\right)$$$.
$$$f{\left(8,32 \right)} = 64$$$
Take the point $$$\left(x, y\right) = \left(\frac{801}{100}, \frac{25600}{801}\right)$$$.
Since $$$f{\left(\frac{801}{100},\frac{25600}{801} \right)} = \frac{1281601}{20025}$$$ is greater than $$$64$$$, it can be stated that $$$64$$$ is the minimum.
Answer
Maximum
No maximum.
Minimum
$$$64$$$A at $$$\left(x, y\right) = \left(8, 32\right)$$$A.