Answer:
Step-by-step explanation:
The method of Lagrange multipliers can be used to find the extrema subject to the constraint. The Lagrangian can be written ...
[tex]L=9x^2+4y^2+3z^2+\lambda(x^2+y^2+z^2-1)[/tex]
We want to find the solution to the simultaneous equations when the partial derivatives are all zero.
[tex]\displaystyle\left\{\begin{array}{l}\dfrac{\partial L}{\partial x}=0=18x+2\lambda x\\\\\dfrac{\partial L}{\partial y}=0=8y+2\lambda y \\\\\dfrac{\partial L}{\partial z}=0=6z+2\lambda z\\\\\dfrac{\partial L}{\partial\lambda}=0=x^2+y^2+z^2-1\end{array}\right.[/tex]
These can be simplified to ...
[tex]\displaystyle\left\{\begin{array}{l}0=x(9+\lambda)\\\\0=y(4+\lambda)\\\\0=z(3+\lambda)\\\\0=x^2+y^2+z^2-1\end{array}\right.[/tex]
The first of these has solutions x=0 or λ=-9. In the latter case, the other equations require y=z=0 and x=1.
The second has solutions y=0 or λ=-4. In the latter case, the other equations require x=z=0 and y=1.
The third has solutions z=0 or λ=-3. In the latter case, the other equations require x=y=0 and z=1.
The objective function (given quadratic form) has these values at the points just found:
9 for (x, y, z) = (1, 0, 0) . . . . . a maximum
4 for (x, y, z) = (0, 1, 0)
3 for (x, y, z) = (0, 0, 1) . . . . . a minimum
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Alternate approach
You can solve the constraint for z^2 and substitute that into the objective function f(x, y, z). It will then be ...
f(x, y) = 9x^2 +4y^2 +3(1 -x^2 -y^2) = 6x^2 +3y^2 +3
Since x^2 and y^2 must be non-negative, the minimum value of this function is clearly 3.
Similarly, you can solve the constraint for x^2 and substitute that into f(x, y, z) to get ...
f(y, z) = 9(1 -y^2 -z^2) +4y^2 +3z^2 = -5y^2 -6z^2 +9
Again, the fact that y^2 and z^2 are zero at least means the maximum value of f(y, z) is 9.
