The equation of the unit sphere, S, centered at the origin is

x^{2} + y^{2 }+ z^{2 }= 1 (Eq. 1)

In particular z^{2 }=1-(x^{2} + y^{2 }) for every point (x,y,z) on S,

The restriction of T(x,y,z) is then

T(x,y,z)=f(x,y)=200xy(1-x^{2} - y^{2})

The partial derivatives of f are

f_{x}(x,y)=200y(1-3x^{2} - y^{2})

f_{y}(x,y)=200x(1-3y^{2} - x^{2})

At a critical point, we have

f_{x}(x,y) = 0 = fy(x,y) (C)

Solving the system (C)

f_{x}(x,y) = 0 iff y = 0 or 3x^{2} + y^{2} =1

and

fy(x,y)=0 iff x=0 or x^{2} + 3y^{2} =1

We have the following:

(I) x = y =0,

or

(II) x = 0 and 3x^{2} + y^{2} =1,

or

(III) y = 0 and x^{2} + 3y^{2} =1,

or

(IV) 3x^{2} + y^{2} =1 and x^{2} + 3y^{2} =1.

Case I: x = y =0 implies that z = ±1 and

T(0,0,±1)=0.

Case II: x = 0 and 3x^{2} + y^{2} =1 implies that y = ±1 and z = 0. The temperature is T(0,±1,0)=0.

Case III: y = 0 and x^{2} + 3y^{2} =1 implies that x = ±1 and z = 0. The temperature is T(±1,0,0)=0.

Case IV: 3x^{2} + y^{2} =1 and x^{2} + 3y^{2} =1 implies that

x =±1/2 , y = ±1/2, z= ±1/√2. We have the following points with their corresponding temperatures

T(1/2,1/2,±1/√2) =T(-1/2,-1/2,±1/√2)= 25

T(-1/2,1/2,±1/√2) = T(1/2,-1/2,±1/√2) = -25.

Hence, the highest temperature is 25.