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Alik [6]
3 years ago
12

This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex

treme values of the function subject to the given constraint. f(x, y, z) = 8x + 8y + 4z; 4x2 + 4y2 + 4z2 = 36
Mathematics
1 answer:
harina [27]3 years ago
5 0
L(x,y,z,\lambda)=8x+8y+4z+\lambda(4x^2+4y^2+4z^2-36)

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

yL_x=y+\lambda xy=0
xL_y=x+\lambda xy=0
\implies yL_x-xL_y=y-x=0\implies y=x

2zL_x=2z+2\lambda xz=0
xL_z=x+2\lambda xz=0
\implies 2zL_x-xL_z=2z-x=0\implies x=2z

2zL_y=2z+2\lambda yz=0
yL_z=y+2\lambda yz=0
\implies 2zL_y-yL_z=2z-y=0\implies y=2z

x=y=2z\implies x^2+y^2+z^2=9\iff 4z^2+4z^2+z^2=9z^2=9\implies z=\pm1

z=\pm1\implies y=x=\pm2

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.
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