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Lagrange Multiplier: Maximum Temperature T = 400xyz² on the Unit Sphere
Problem. The temperature at any point \((x,y,z)\) in space is \(T=400xyz^2\). Find the maximum temperature on the surface of the unit sphere \(x^2+y^2+z^2=1\).
Video explanation
Step 1: Auxiliary function
Constraint: \(x^2+y^2+z^2-1=0\).
$$F(x,y,z,\lambda)=400xyz^2+\lambda\,(x^2+y^2+z^2-1)$$
Step 2: Partial derivatives
$$F_x=400yz^2+2\lambda x,\quad F_y=400xz^2+2\lambda y,\quad F_z=800xyz+2\lambda z$$
Step 3: Stationary points
Set \(F_x=F_y=F_z=0\) and solve each for \(-\lambda\):
$$-\lambda=\frac{200yz^2}{x}\ \ (1),\qquad -\lambda=\frac{200xz^2}{y}\ \ (2),\qquad -\lambda=400xy\ \ (3)$$
From (1) and (2):
$$\frac{y}{x}=\frac{x}{y}\;\Rightarrow\;y^2=x^2$$
From (2) and (3):
$$\frac{200xz^2}{y}=400xy\;\Rightarrow\;z^2=2y^2\;\Rightarrow\;y^2=\frac{z^2}{2}$$
So \(x^2=y^2=\dfrac{z^2}{2}\). Substitute into the constraint:
$$\frac{z^2}{2}+\frac{z^2}{2}+z^2=1\;\Rightarrow\;2z^2=1\;\Rightarrow\;z^2=\frac12$$
$$x^2=y^2=\frac14\;\Rightarrow\;x=\pm\frac12,\quad y=\pm\frac12,\quad z=\pm\frac1{\sqrt2}$$
Step 4: Evaluate T
Since \(T=400\,xy\,z^2\) and \(z^2=\tfrac12\) in every case, the sign of \(T\) depends only on the sign of \(xy\):
| x | y | z | T |
|---|---|---|---|
| 1/2 | 1/2 | ±1/√2 | 50 |
| −1/2 | −1/2 | ±1/√2 | 50 |
| 1/2 | −1/2 | ±1/√2 | −50 |
| −1/2 | 1/2 | ±1/√2 | −50 |
$$T=400\cdot\frac12\cdot\frac12\cdot\left(\frac1{\sqrt2}\right)^2=400\cdot\frac14\cdot\frac12=50$$
Final answer
Maximum temperature50at \(\left(\tfrac12,\tfrac12,\pm\tfrac1{\sqrt2}\right)\) and \(\left(-\tfrac12,-\tfrac12,\pm\tfrac1{\sqrt2}\right)\)
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