Question

Use Lagrange multipliers to find the maximum and minimum values of subject to the given constraint. Also, find the points at which these extreme values occur.
f(x, y) xy: 48x+3y2 3456
Enter your answers for the points in order of increasing x-value.
Maximum______
Minimum_______

Answer 1

Answer:

Following are the responses to the given question:

Step-by-step explanation:

Given:

$\to f(x,y)=xy\\\\\to 48x^2+3y^2=3456$

let $\phi(x,y)=48x^2+3y^2=3456$

using the lagrange multiplies:

$\bigtriangledown f=\lambda \bigtriangledown \phi\\\\i)\ \ y=96 \lambda x \to \lambda=\frac{y}{96\ x} \\\\ii)\ \ x=6 \lambda y \to \lambda=\frac{y}{6\ y} \\\\iii)\ \ 48x^2+3y^2=3456$

From equation (i) and (ii)\\\\

$\to \frac{y}{96x}=\frac{x}{6y}\\\\\to 6y^2=96x^2\\\\\to 3y^2=48x^2$

From equation 3:

$48x^2+48x^2=3456\\\\96x^2 3456\\\\x^2=36\\\\x=\pm 6\\\\\therefore 3y^2=48(36)\\\\3y^2=1728\\\\y^2=576\\\\y=\pm24\\\\(x,y)=(\pm 6,\pm 24)\\\\Maximum \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 144 \ \ \ \ at \ \ (6,24) \ \ \ \ \ \ \ and \ \ \ \ \ \ (-6, -24) \\\\Minimum \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -144 \ \ \ \ at \ \ (-6,24) \ \ \ \ \ \ \ and \ \ \ \ \ \ (6,-24)$