Question

What is the max. value of T(x,y,z) = 1+xy+yz given that x,y and z all lie in a unit sphere?

Answer 1

pkay then, let the unit sphere be: x^2 + y^2 +z^2 =1

now you can maximize 1+xy+yz for the above constraint now.

Answer 2

Use spherical coordinates as in
http://tutorial.math.lamar.edu/Classes/CalcII/SphericalCoords.aspx
Replace x, y, z by their values. Since we are on the unit sphere
\[ \rho =1\]
The function to be maximized becomes
\[ h(\theta,\phi)= \sin (\theta ) \cos (\theta )
\sin ^2(\phi )+\sin (\theta )
\sin (\phi ) \cos (\phi )+1
\]
It is gradient is
\[
gradienth(\theta,\phi)=\left\{-\sin ^2(\theta ) \sin
^2(\phi )+\cos ^2(\theta )
\sin ^2(\phi )+\cos (\theta )
\sin (\phi ) \cos (\phi
),-\sin (\theta ) \sin
^2(\phi )+\sin (\theta ) \cos
^2(\phi )+2 \sin (\theta )
\cos (\theta ) \sin (\phi )
\cos (\phi )\right\}
\]
Use your favorite symbolic system to find that the gradient is zero at 15 points. Four of them give a maximum of h \[
1+\frac{1}{\sqrt{2}}
\]

Answer 3

The first point is
\[ \left\{\frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2}\right\}
\]
The second point is

\[

\left\{-\frac{1}{2},-\frac{1}{\sqrt{2}},-\frac{1}{2}\right\}
\]

The last two points are the same of the first two.

Answer 4

Here is how to do it using lagrange multipliers.