91Ó°ÊÓ

The given function is $f(x,y,z)=xyz$and the given constraint is$x^2+4y^2+16z^2=64.$

Let's write the gradient of the given functions,

$$\begin{gathered} \mathrm{∇}f(x,y,z)=yz\mathrm{i}+yz\mathrm{j}+xy\mathrm{k} \\ \mathrm{∇}\mathrm{g}(\mathrm{x},\mathrm{y},\mathrm{z})=2\mathrm{xi}+8\mathrm{yj}+32\mathrm{zk} \end{gathered}$$

Use the Lagrange's method, so the equation is,

role="math" localid="1649931058104" $$\begin{gathered} \mathrm{∇}f(x,y,z)=\mathrm{λ}\mathrm{∇}g(x,y,z) \\ yz\mathbf{i}+yz\mathbf{j}+xy\mathbf{k}=\mathrm{λ}(2x\mathbf{i}+8y\mathbf{j}+32z\mathbf{k}) \\ yz\mathbf{i}+yz\mathbf{j}+xy\mathbf{k}=2\mathrm{λ}x\mathbf{i}+8\mathrm{λ}y\mathbf{j}+32\mathrm{λ}z\mathbf{k} \end{gathered}$$

Here, $yz=2\mathrm{λ}x,xz=8\mathrm{λ}y,\text{and}xy=32\mathrm{λ}z.$

The values of $\mathrm{λ},$we get are $\mathrm{λ}=\frac{yz}{2x}=\frac{xz}{8y}=\frac{xy}{32z}.$

Thus, $x^2=4y^2\text{and}{z}^2=\frac{y^2}{4}.$

Substitute the above value in the given constraint,

$$\begin{gathered} x^2+4y^2+16z^2=64 \\ \begin{array}{rl}4y^2+4y^2+4y^2& =64\end{array} \\ 12y^2=64 \\ y=Â\pm \frac{4}{\sqrt{3}} \end{gathered}$$

So, $x=Â\pm \frac{8}{\sqrt{3}},\text{and}z=Â\pm \frac{2}{\sqrt{3}}.$

Hence the points we get arerole="math" localid="1649931596298" $$\begin{gathered} (−\frac{8}{\sqrt{3}},−\frac{4}{\sqrt{3}},−\frac{2}{\sqrt{3}}),(\frac{8}{\sqrt{3}},−\frac{4}{\sqrt{3}},−\frac{2}{\sqrt{3}}),(\frac{8}{\sqrt{3}},−\frac{4}{\sqrt{3}},\frac{2}{\sqrt{3}}), \\ (−\frac{8}{\sqrt{3}},−\frac{4}{\sqrt{3}},\frac{2}{\sqrt{3}}),(−\frac{8}{\sqrt{3}},\frac{4}{\sqrt{3}},−\frac{2}{\sqrt{3}}),(\frac{8}{\sqrt{3}},\frac{4}{\sqrt{3}},−\frac{2}{\sqrt{3}}), \\ (\frac{{\displaystyle 8}}{{\displaystyle \sqrt{3}}},\frac{{\displaystyle 4}}{{\displaystyle \sqrt{3}}},\frac{{\displaystyle 2}}{{\displaystyle \sqrt{3}}}),\text{and}(−\frac{8}{\sqrt{3}},\frac{4}{\sqrt{3}},\frac{2}{\sqrt{3}}). \end{gathered}$$

Now, let's find the value of the function through the points.

So,

When the points are $(−\frac{8}{\sqrt{3}},−\frac{4}{\sqrt{3}},−\frac{2}{\sqrt{3}}),(\frac{8}{\sqrt{3}},−\frac{4}{\sqrt{3}},\frac{2}{\sqrt{3}}),(\frac{8}{\sqrt{3}},\frac{4}{\sqrt{3}},−\frac{2}{\sqrt{3}})$

$\text{and}(−\frac{8}{\sqrt{3}},\frac{4}{\sqrt{3}},\frac{2}{\sqrt{3}})$the value of the function is $-\frac{64}{3\sqrt{3}}.$

And when the points are $(\frac{8}{\sqrt{3}},−\frac{4}{\sqrt{3}},−\frac{2}{\sqrt{3}}),(−\frac{8}{\sqrt{3}},−\frac{4}{\sqrt{3}},\frac{2}{\sqrt{3}}),(−\frac{8}{\sqrt{3}},\frac{4}{\sqrt{3}},−\frac{2}{\sqrt{3}}),$

$\mathrm{and}(\frac{8}{\sqrt{3}},\frac{4}{\sqrt{3}},\frac{2}{\sqrt{3}})$the value of the function is $\frac{64}{3\sqrt{3}}.$

The maximum value of the function is $\frac{64}{3\sqrt{3}}$and the minimum value of the function is $-\frac{64}{3\sqrt{3}}.$

As the constraint is an ellipsoid that is bounded and closed, so maximum and minimum both exist.