In Exercises, find the maximum and minimum of the function f subject to the given constraint. In each case explain why the maximum and minimum must both exist.

$f(x,y,z)=x+y+z\text{when}{x}^2+y^2+z^2=9$

Given function is$f(x,y,z)=x+y+z.$

Given constraint is$x^2+y^2+z^2=9.$

Gradients of function.

$$\begin{gathered} \nabla f(x,y,z)=i+j+k \\ \nabla g(x,y,z)=2xi+2yj+2zk \end{gathered}$$

Use the method of Lagrange multipliers.

$$\begin{gathered} \nabla f(x,y,z)=\lambda \nabla g(x,y,z) \\ i+j+k=\lambda \left(2xi+2yj+2zk\right) \\ i+j+k=2\lambda xi+2\lambda yj+2\lambda zk \end{gathered}$$

Compare terms.

$$\begin{gathered} 1=2\lambda x\Rightarrow \lambda =\frac{1}{2x} \\ 1=2\lambda y\Rightarrow \lambda =\frac{1}{2y} \\ 1=2\lambda z\Rightarrow \lambda =\frac{1}{2z} \\ \mathrm{so}x=y=z \end{gathered}$$

substitute $y=x\&z=x$in constraint.

$$\begin{gathered} x^2+y^2+z^2=9 \\ {x}^2+x^2+x^2=9 \\ 3x^2=9 \\ x=\pm \sqrt{3} \\ y=\pm \sqrt{3} \\ z=\pm \sqrt{3} \end{gathered}$$

so critical points are$\left(-\sqrt{3},-\sqrt{3},-\sqrt{3}\right)\&\left(\sqrt{3},\sqrt{3},\sqrt{3}\right).$

Find function value at $\left(-\sqrt{3},-\sqrt{3},-\sqrt{3}\right).$

$$\begin{gathered} f(x,y,z)=x+y+z \\ f(-\sqrt{3},-\sqrt{3},-\sqrt{3})=\left(-\sqrt{3}\right)+\left(-\sqrt{3}\right)+\left(-\sqrt{3}\right) \\ f(-\sqrt{3},-\sqrt{3},-\sqrt{3})=-3\sqrt{3} \end{gathered}$$

Find the function value at $\left(\sqrt{3},\sqrt{3},\sqrt{3}\right).$

role="math" localid="1649895024258" $$\begin{gathered} f(x,y,z)=x+y+z \\ f(\sqrt{3},\sqrt{3},\sqrt{3})=\sqrt{3}+\sqrt{3}+\sqrt{3} \\ f(\sqrt{3},\sqrt{3},\sqrt{3})=3\sqrt{3} \end{gathered}$$

So the maximum value of the function is $3\sqrt{3}$and the minimum value is $3\sqrt{3}.$

As constraint is bounded and closed sphere of radius $3$so maximum and minimum both exist.