Respuesta :
Check for critical points within the unit ball by solving for when the first-order partial derivatives vanish:
[tex]f_x=5yz=0\implies y=0\text{ or }z=0[/tex]
[tex]f_y=5xz=0\implies x=0\text{ or }z=0[/tex]
[tex]f_z=5xy=0\implies x=0\text{ or }y=0[/tex]
Taken together, we find that (0, 0, 0) appears to be the only critical point on [tex]f[/tex] within the ball. At this point, we have [tex]f(0,0,0)=0[/tex].
Now let's use the method of Lagrange multipliers to look for critical points on the boundary. We have the Lagrangian
[tex]L(x,y,z,\lambda)=5xyz+\lambda(x^2+y^2+z^2-1)[/tex]
with partial derivatives (set to 0)
[tex]L_x=5yz+2\lambda x=0[/tex]
[tex]L_y=5xz+2\lambda y=0[/tex]
[tex]L_z=5xy+2\lambda z=0[/tex]
[tex]L_\lambda=x^2+y^2+z^2-1=0[/tex]
We then observe that
[tex]xL_x+yL_y+zL_z=0\implies15xyz+2\lambda=0\implies\lambda=-\dfrac{15xyz}2[/tex]
So, ignoring the critical point we've already found at (0, 0, 0),
[tex]5yz+2\left(-\dfrac{15xyz}2\right)x=0\implies5yz(1-3x^2)=0\implies x=\pm\dfrac1{\sqrt3}[/tex]
[tex]5xz+2\left(-\dfrac{15xyz}2\right)y=0\implies5xz(1-3y^2)=0\implies y=\pm\dfrac1{\sqrt3}[/tex]
[tex]5xy+2\left(-\dfrac{15xyz}2\right)z=0\implies5xy(1-3z^2)=0\implies z=\pm\dfrac1{\sqrt3}[/tex]
So ultimately, we have 9 critical points - 1 at the origin (0, 0, 0), and 8 at the various combinations of [tex]\left(\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3}\right)[/tex], at which points we get a value of either of [tex]\pm\dfrac5{\sqrt3}[/tex], with the maximum being the positive value and the minimum being the negative one.
[tex]f_x=5yz=0\implies y=0\text{ or }z=0[/tex]
[tex]f_y=5xz=0\implies x=0\text{ or }z=0[/tex]
[tex]f_z=5xy=0\implies x=0\text{ or }y=0[/tex]
Taken together, we find that (0, 0, 0) appears to be the only critical point on [tex]f[/tex] within the ball. At this point, we have [tex]f(0,0,0)=0[/tex].
Now let's use the method of Lagrange multipliers to look for critical points on the boundary. We have the Lagrangian
[tex]L(x,y,z,\lambda)=5xyz+\lambda(x^2+y^2+z^2-1)[/tex]
with partial derivatives (set to 0)
[tex]L_x=5yz+2\lambda x=0[/tex]
[tex]L_y=5xz+2\lambda y=0[/tex]
[tex]L_z=5xy+2\lambda z=0[/tex]
[tex]L_\lambda=x^2+y^2+z^2-1=0[/tex]
We then observe that
[tex]xL_x+yL_y+zL_z=0\implies15xyz+2\lambda=0\implies\lambda=-\dfrac{15xyz}2[/tex]
So, ignoring the critical point we've already found at (0, 0, 0),
[tex]5yz+2\left(-\dfrac{15xyz}2\right)x=0\implies5yz(1-3x^2)=0\implies x=\pm\dfrac1{\sqrt3}[/tex]
[tex]5xz+2\left(-\dfrac{15xyz}2\right)y=0\implies5xz(1-3y^2)=0\implies y=\pm\dfrac1{\sqrt3}[/tex]
[tex]5xy+2\left(-\dfrac{15xyz}2\right)z=0\implies5xy(1-3z^2)=0\implies z=\pm\dfrac1{\sqrt3}[/tex]
So ultimately, we have 9 critical points - 1 at the origin (0, 0, 0), and 8 at the various combinations of [tex]\left(\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3}\right)[/tex], at which points we get a value of either of [tex]\pm\dfrac5{\sqrt3}[/tex], with the maximum being the positive value and the minimum being the negative one.
The maximum value of the given function is [tex]\frac{5}{3\sqrt{3} }[/tex]
And the minimum value is [tex]\frac{-5}{3\sqrt{3} }[/tex]
Given:
[tex]f(x,y,z)=5xyz[/tex]
Computation:
Differentiating the given equation up to second order
[tex]f_x=5yz\Rightarrow 5yz=0 either y=0 or z=0\\f_y=0\Rightarrow 5xz=0 either x=0 or z=0\\f_z=0\Rightarrow 5xy=0 either x=0 or y=0[/tex]
So, the critical point is (0,0,0)
Now, using the Lagrange's on the boundary,
[tex]g(x,y,z)=x^2+y^2+z^2-1=0\\g_x=2x\\g_y=2y\\g_z=2z[/tex]
So, [tex]\bigtriangledown f=\lambda \bigtriangledown g\left<5yz,5xz,5xy \right>=\lambda \left<2x,2y,2z \right>[/tex]
By solving we get,
[tex]x^2=y^2=z^2[/tex] then,
[tex]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} } \\[/tex]
So, the critical points are (0,0,0),[tex](\frac{1}{\sqrt{3} } ,\frac{1}{\sqrt{3} } ,\frac{1}{\sqrt{3} } )[/tex] and [tex](\frac{-1}{\sqrt{3} }, \frac{-1}{\sqrt{3} },\frac{-1}{\sqrt{3} })[/tex]
So, by substituting the critical points we get,
[tex]f(0,0,0)=0\\f(\frac{1}{\sqrt{3} }, \frac{1}{\sqrt{3} },\frac{1}{\sqrt{3} })=5(\frac{1}{\sqrt{3} })(\frac{1}{\sqrt{3} })(\frac{1}{\sqrt{3} })\\=\frac{5}{3\sqrt{3} } \\f(-\frac{1}{\sqrt{3} },-\frac{1}{\sqrt{3} },-\frac{1}{\sqrt{3} })=5(-\frac{1}{\sqrt{3} })(-\frac{1}{\sqrt{3} })(-\frac{1}{\sqrt{3} })\\=-\frac{5}{3\sqrt{3} }[/tex]
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