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Jobisdone [24]
3 years ago
9

Find the max and min values of f(x,y,z)=x+y-z on the sphere x^2+y^2+z^2=81

Mathematics
1 answer:
Anton [14]3 years ago
5 0
Using Lagrange multipliers, we have the Lagrangian

L(x,y,z,\lambda)=x+y-z+\lambda(x^2+y^2+z^2-81)

with partial derivatives (set equal to 0)

L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}
L_y=1+2\lambda y=0\implies y=-\dfrac1{2\lambda}
L_z=-1+2\lambda z=0\implies z=\dfrac1{2\lambda}
L_\lambda=x^2+y^2+z^2-81=0\implies x^2+y^2+z^2=81

Substituting the first three equations into the fourth allows us to solve for \lambda:

x^2+y^2+z^2=\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}=81\implies\lambda=\pm\dfrac1{6\sqrt3}

For each possible value of \lambda, we get two corresponding critical points at (\mp3\sqrt3,\mp3\sqrt3,\pm3\sqrt3).

At these points, respectively, we get a maximum value of f(3\sqrt3,3\sqrt3,-3\sqrt3)=9\sqrt3 and a minimum value of f(-3\sqrt3,-3\sqrt3,3\sqrt3)=-9\sqrt3.
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