Question

The function ​f(x,y,z)equals2 x plus z squared has an absolute maximum value and absolute minimum value subject to the constraint x squared plus 2 y squared plus 3 z squaredequals16. Use Lagrange multipliers to find these values.

Answer 1

Answer:

Absolute maxima an minma both occured at $\frac{25}{3}$.

Step-by-step explanation:

Given function is,

$f(x,y,z)=2x+z^2\hfill (1)$

subject to,

$x^2+2y^2+3z^2=16\hfill (2)$

Let $g(x,y,z)=x^2+2y^2=3z^2-16$

To find absolute maxima and absolute minima using Lagranges multipliers method consider $\lambda$ as the multipliers such that,

$\nabla f=\lambda \nabla g$

$\leftrightarrow (2, 0 ,2z )=\lambda (2x, 4y, 6z)$

on compairing both side we get,

$2z=6\lambda z\implies \lambda=\frac{1}{3}$

$4\labda y=0\implies y=0$

$2=2\lambda x\implies x=\frac{1}{\lambda}=3$

From (2),

$x^2+2y^2+3z^2=16$

$\implies 9+0+3z^2=16$

$\implies z=\pm\sqrt{\frac{7}{3}}$

Absolute maxima, at x=3, y=0,$z= \sqrt{\frac{7}{3}}$ is,

$|f(x,y,z)|_{max}=(2x+z^2)_(3,0,\sqrt{\frac{7}{3}})=(2\times3)+\frac{7}{3}=\frac{25}{3}$

Absolute minima, at x=3, y=0, $z= -\sqrt{\frac{7}{3}}$ is,

$|f(x,y,z)|_{max}=(2x+z^2)_(3,0,-\sqrt{\frac{7}{3}})=(2\times3)+\frac{7}{3}=\frac{25}{3}$

Hence the result.