Respuesta :
Answer:
Absolute maxima an minma both occured at [tex]\frac{25}{3}[/tex].
Step-by-step explanation:
Given function is,
[tex]f(x,y,z)=2x+z^2\hfill (1)[/tex]
subject to,
[tex]x^2+2y^2+3z^2=16\hfill (2)[/tex]
Let [tex]g(x,y,z)=x^2+2y^2=3z^2-16[/tex]
To find absolute maxima and absolute minima using Lagranges multipliers method consider [tex]\lambda[/tex] as the multipliers such that,
[tex]\nabla f=\lambda \nabla g[/tex]
[tex]\leftrightarrow (2, 0 ,2z )=\lambda (2x, 4y, 6z)[/tex]
on compairing both side we get,
[tex]2z=6\lambda z\implies \lambda=\frac{1}{3}[/tex]
[tex]4\labda y=0\implies y=0[/tex]
[tex]2=2\lambda x\implies x=\frac{1}{\lambda}=3[/tex]
From (2),
[tex]x^2+2y^2+3z^2=16[/tex]
[tex]\implies 9+0+3z^2=16[/tex]
[tex]\implies z=\pm\sqrt{\frac{7}{3}}[/tex]
Absolute maxima, at x=3, y=0,[tex] z= \sqrt{\frac{7}{3}}[/tex] is,
[tex]|f(x,y,z)|_{max}=(2x+z^2)_(3,0,\sqrt{\frac{7}{3}})=(2\times3)+\frac{7}{3}=\frac{25}{3}[/tex]
Absolute minima, at x=3, y=0, [tex]z= -\sqrt{\frac{7}{3}}[/tex] is,
[tex]|f(x,y,z)|_{max}=(2x+z^2)_(3,0,-\sqrt{\frac{7}{3}})=(2\times3)+\frac{7}{3}=\frac{25}{3}[/tex]
Hence the result.
