Solutions of the test on partial derivatives

The domain is the exterior of the unit circle centered at the origin including the circle itself. The range is the set of numbers greater than or equal to 0. The level curve is the circle of radius 2 centered at the origin.

The equation of the tangent plane is

        g(x,y,z)=100+(x-300)/9+(y-300)/9+(z-300)/9

The estimate for w is

        g(301,299,302)=100.22...

By the chain rule,

        g_s(1,1)=3*2+7*4+(-2)*5=24

and

        g_t(1,1)=3*(-2)+7*(-4)+(-2)*0=-34

The gradient of T at P is
```
        -4i-2j-4k
```
The rate of change of T at P in the direction of u is the dot product of the gradient vector and u. Hence it is
```
        D_uT(2,1,1)=(-4)*(2/3)+(-2)*(1/3)+(-4)*(-2/3)=-2/3
```
The direction to cool off most rapidly is opposite to the direction of the gradient. The corresponding unit vector is
```
        (2/3)i+(1/3)j+(2/3)k
```
The maximum rate of change of T at P is the length of the gradient vector. Thus it is given by
```
        (16+4+16)^1/2=6
```
The equation of the tangent plane is
```
        -x+3y+2z=18
```
There are three critical points
```
        P₁(0,0), P_2,3(-1/3,±2^1/2/3)
```
By the second derivative test, we find that P₁ is a local minimum and the other two critical points are saddle points.
We find five critical points with the following function values
```
        f(2/5,1/5)=6/5, f(1/2,0)=5/4, f(0,0)=1, f(1,0)=1, f(0,1)=2
```
Therefore, the absolute maximum value is 1 and the absolute minimum value is 0.
We find four critical points with the following functions values
```
        f(0,1)=-2, f(0,-1)=0, f(±15^1/2/4,-1/4)=9/8
```
Therefore, the absolute maximum value is 9/8 and the absolute minimum value is -2.