contestada

This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the extreme values of the function subject to the given constraint. f(x, y, z) = 8x + 8y + 4z; 4x2 + 4y2 + 4z2 = 36

Respuesta :

LammettHash
[tex]L(x,y,z,\lambda)=8x+8y+4z+\lambda(4x^2+4y^2+4z^2-36)[/tex]

[tex]L_x=8+8\lambda x=0\implies 1+\lambda x=0[/tex]
[tex]L_y=8+8\lambda y=0\implies 1+\lambda y=0[/tex]
[tex]L_z=4+8\lambda z=0\implies 1+2\lambda z=0[/tex]
[tex]L_\lambda=4x^2+4y^2+4z^2-36=0\implies x^2+y^2+z^2=9[/tex]

[tex]yL_x=y+\lambda xy=0[/tex]
[tex]xL_y=x+\lambda xy=0[/tex]
[tex]\implies yL_x-xL_y=y-x=0\implies y=x[/tex]

[tex]2zL_x=2z+2\lambda xz=0[/tex]
[tex]xL_z=x+2\lambda xz=0[/tex]
[tex]\implies 2zL_x-xL_z=2z-x=0\implies x=2z[/tex]

[tex]2zL_y=2z+2\lambda yz=0[/tex]
[tex]yL_z=y+2\lambda yz=0[/tex]
[tex]\implies 2zL_y-yL_z=2z-y=0\implies y=2z[/tex]

[tex]x=y=2z\implies x^2+y^2+z^2=9\iff 4z^2+4z^2+z^2=9z^2=9\implies z=\pm1[/tex]

[tex]z=\pm1\implies y=x=\pm2[/tex]

So we have two critical points, (2, 2, 1) and (-2, -2, -1), which respectively give a max of 36 and a min of -36.
Cricetus

The extreme max value is "[tex]f(x,y,z) = 36[/tex]" and min value is "[tex]f(x,y,z) = -36[/tex]".

According to the question,

  • [tex]f(x, y, z) = 8x + 8y + 4z[/tex]

let,

  • [tex]g(x,y,z) = 4x^2 + 4y^2 + 4z^2 - 36[/tex]

By using Lagrange multipliers,

→ [tex]\bigtriangledown f= \lambda \bigtriangledown g[/tex]...(equation 1)

  • [tex]\bigtriangledown f = <8,8,4>[/tex]
  • [tex]\bigtriangledown g = <8x, 8y,8z>[/tex]

∴ [tex]<8,8,4>=<8\lambda x, 8 \lambda y, 8 \lambda z >[/tex]

then,

  • [tex]8 \lambda x = 8[/tex]...(equation 2)
  • [tex]8 \lambda y =8[/tex]...(equation 3)
  • [tex]8 \lambda z = 4[/tex]...(equation 4)

→ [tex](x = \frac{1}{\lambda}, y=\frac{1}{\lambda}, z =\frac{1}{2 \lambda} )[/tex]

As we know,

→             [tex]4x^2 +4y^2+4z^2=36[/tex]

By substituting the values, we get

→ [tex]4(\frac{1}{\lambda} )^2+ 4(\frac{1}{\lambda} )^2+4(\frac{1}{2\lambda} )^2 =36[/tex]

→                 [tex]\frac{4}{\lambda^2} +\frac{4}{\lambda^2} +\frac{1}{\lambda^2} =36[/tex]

→                                  [tex]\frac{9}{\lambda^2}=36[/tex]

                                     [tex]\frac{1}{\lambda}= \pm \frac{6}{3}[/tex]

                                     [tex]\frac{1}{\lambda} = \pm 2[/tex]

∴ x = 2, y = 2, z = 1

then,

→ [tex]f(x,y,z) = 36[/tex] (Extreme max value)

and,

∴ x = -2, y = -2, z = -1

then,

→ [tex]f(x,y,z) = -36[/tex] (min value)

Thus the above answer is correct.

Learn more about Lagrange multipliers here:

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