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Sagot :
the maximum and minimum values of f(x,y,z) = 2xyz are [tex]\frac{2}{\sqrt{3} } and \frac{-2}{\sqrt{3} }[/tex]
f(x,y,z)=2xyz
Computation:
Differentiating the given equation up to second order
[tex]\begin{gathered}f_x=2yz\Rightarrow 2yz=0 either y=0 or z=0\\f_y=0\Rightarrow 2xz=0 either x=0 or z=0\\f_z=0\Rightarrow 2xy=0 either x=0 or y=0\end{gathered}[/tex]
So, the critical point is (0,0,0)
Now, using the Lagrange's on the boundary,
[tex]\begin{gathered}g(x,y,z)=x^2+y^2+z^2-1=0\\g_x=2x\\g_y=2y\\g_z=2z\end{gathered}[/tex]
[tex]So, \bigtriangledown f=\lambda \bigtriangledown g\left < 5yz,5xz,5xy \right > =\lambda \left < 2x,2y,2z )[/tex]
By solving we get,
[tex]x^2=y^2=z^2[/tex] then,
[tex]\begin{gathered}x^2+y^2+z^2=1\\3z^2=1\\z=\pm \frac{1}{\sqrt{3} } \\y=\pm \frac{1}{\sqrt{3} } \\\\x=\pm \frac{1}{\sqrt{3} } \\\end{gathered} x 2+y 2+z 2=13z 2=1z=± 31[/tex]
[tex]So, the critical points are (0,0,0),(\frac{1}{\sqrt{3} } ,\frac{1}{\sqrt{3} } ,\frac{1}{\sqrt{3} } )and (\frac{-1}{\sqrt{3} }, \frac{-1}{\sqrt{3} },\frac{-1}{\sqrt{3} })[/tex]
So, by substituting the critical points we get,
[tex]\begin{gathered}f(0,0,0)=0\\f(\frac{1}{\sqrt{3} }, \frac{1}{\sqrt{3} },\frac{1}{\sqrt{3} })=2(\frac{1}{\sqrt{3} })(\frac{1}{\sqrt{3} })(\frac{1}{\sqrt{3} })\\=\frac{2}{3\sqrt{3} } \\f(-\frac{1}{\sqrt{3} },-\frac{1}{\sqrt{3} },-\frac{1}{\sqrt{3} })=2(-\frac{1}{\sqrt{3} })(-\frac{1}{\sqrt{3} })(-\frac{1}{\sqrt{3} })\\=-\frac{2}{3\sqrt{3} }\end{gathered}[/tex]
Hence the maximum and minimum values of f(x,y,z) are [tex]\frac{2}{\sqrt{3} } and \frac{-2}{\sqrt{3} }[/tex]
Learn more about maximum and minimum values here https://brainly.com/question/4609414
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