Finding the maxima of a 3 variable function - basics

Question

I am stuck on this problem and unfortunately the reading for my class does not include any example for finding a maxima of a 3 variable function.

$T(x,y,z) = -(x^2+y^2+z^2)+4x+8y+2z+11$ representing a temperature of a room.

I need to find the point $(x,y,z)$ where the temperature is at the maximum, and what is the value of this maximum.

Usually, for a two variable function, I would find the critical points of the function and see if the conditions for a maximum apply.

But for the function above, I do not know how to even start the problem. Any help would be appreciated.

Thank you.

Answer 1

You need to find partial derivatives and solve for partial derivatives equal zero. $$T(x,y,z) = -(x^2+y^2+z^2)+4x+8y+2z+11$$

$$T_x = -2x+4=0 \implies x=2$$ $$T_y = -2y+8=0 \implies y=4$$ $$T_z = -2z+2=0 \implies z=1$$

Since $$T_{xx}T_{yy}-T_{xy}T_{xy}=4>0$$

And $T_{xx}=-2<0$, the point $(2,4,1)$ is a maximizer.

the maximum value is $T(2,4,1) =32$

Answer 2

Guide:

Same procedure.

Find $\frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z}$, equate them to zero and solve for the stationary point.

The hessian for this problem is $-2I$ which is negative definite, the stationary point is a minimal point.

Answer 3

Let us complete the squares.

$$T(x,y,z) = -(x^2+y^2+z^2)+4x+8y+2z+11 = $$

$$32-(x-2)^2 - (y-4)^2-(z-1)^2 \le 32$$

As you see the maximum value of $32$ is attained at $(2,4,1)$