Lagrange multiplier - space probe

Question

i am stuck on this question which uses the Lagrange multiplier. I am trying to construct the equations using the partial derivatives but the $x$'s and $y$'s cancel. can anyone help?

A space probe in the shape of the ellipsoid

$x^2 + y^2 + 3z^2 = 3$

enters a planet's atmosphere and begins to heat up. The temperature on its surface is found to be $T(x, y, z) = x^2 + 2y^2 + 6z$:

Use the method of Lagrange multipliers to find the hottest points on the probe's surface.

Answer 1

Taking the gradient of $T$, $$\nabla T=(T_x,T_y,T_z)=(2x,4y,6)$$ Taking the gradient of $g$ times $\lambda$, $$\lambda\nabla g=(\lambda g_x,\lambda g_y,\lambda g_z)=(\lambda 2x,\lambda 2,\lambda g_x)$$ Setting them equal $$2x=\lambda 2x \Rightarrow 2x(1-\lambda)=0$$ $$4y=\lambda 2y \Rightarrow 2y(2-\lambda)=0$$ $$6=\lambda 6z \Rightarrow \lambda=\frac1{z}$$ So what if $\lambda=1$? Then from the second equation, $2y=0$ and thus $y=0$. Also, fron the third equation, we have $1=\frac1{z}$ and thus, $z=1$. Plugging into the constraint equation, we have $$x^2+0^2+3(1)^2=3$$ $$x^2+3=3$$ $$x^2=0$$ $$x=0$$ Thus the critical point when $\lambda =1$ is $(0,0,1)$.

What if $\lambda=2$? Then from the first equation, $2x(-1)=-2x=0$ and thus $x=0$. Also, from the third equation, we have $2=\frac1{z}$ and thus $z=\frac1{2}$. Plugging into the constraint equation we have $$0^2+y^2+3\left(\frac1{2}\right)^2=3$$ $$y^2+\frac3{4}=3$$ $$y^2=\frac9{4}$$ $$y=\pm\frac3{2}$$ Thus the critical points when $\lambda =2$ is $(0,\pm\frac3{2},\frac1{2})$.

In the case that $x=0$ and $y=0$, we can simply plug $x$ and $y$ into the constraint equation and solve for $z$ $$0^2+0^2+3z^2=3$$ $$3z^2=3$$ $$z^2=1$$ $$z=\pm1$$ Thus the critical points when $x=0,y=0$ is $(0,0,\pm1)$.

Finally, plugging the critical points into $T$ $$T(0,0,1)=0^2+2\cdot0^2+6\cdot 1=6$$ $$T(0,0,-1)=0^2+2\cdot0^2+6\cdot(-1)=-6$$ $$T\left(0,\pm\frac3{2},\frac1{2}\right)=0^2+2\cdot\left(\pm\frac3{2}\right)^2+6\left(\frac1{2}\right)=\frac9{2}+3=\frac{15}{2}$$

Answer 2

Set $\Lambda \colon\mathbb R^4\to \mathbb R, (x,y,z,\lambda)\mapsto x^2+2y^2+6z+\lambda (x^2+y^2+3z^2-3)$.

Let $(x,y,z,\lambda)\in \mathbb R^4$.

The following holds:

$$\begin{cases} \Lambda _x(x,y,z,\lambda)&=2x+2\lambda x\\ \Lambda _y(x,y,z,\lambda)&=4y+2\lambda y\\ \Lambda_z(x,y,z, \lambda)&=6+6\lambda _z\\ \Lambda _\lambda(x,y,z,\lambda)&=x^2+y^2+3z^2-3.\end{cases}$$

Suppose $$\begin{cases} 0&=2x+2\lambda x\\ 0&=4y+2\lambda y\\ 0&=6+6\lambda z\\ 0&=x^2+y^2+3z^2-3,\end{cases}$$

then $$\begin{cases} 0&=(1+\lambda)x\\ 0&=(2+\lambda)y\\ \lambda&=-\dfrac1 z \land z\neq 0\neq \lambda \\ 0&=x^2+y^2+3z^2-3,\end{cases}$$

which implies $$\begin{cases} \lambda =-1\lor x=0\\ \lambda =-2 \lor y=0\\ \lambda=-\dfrac1 z \land z\neq 0\neq \lambda \\ x^2+y^2+3z^2-3=0.\end{cases}$$

$\boxed{\text{Case }\lambda =-1}$

It follows that $y=0$ and $z=1$. Thus $x^2+0+3-3=0$ and $x=0$, yielding the critical point $\color{blue}{(0,0,1)}$.

$\boxed{\text{Case }x=0}$

• $\boxed{\text{Sub case }\lambda =-2}$ It comes $z=\dfrac 1 2$ and $x^2+y^2-\dfrac 9 4=0$, that is $y=\pm\dfrac 3 2$, yielding the critical points $\color{blue}{\left(0, \dfrac 3 2, \dfrac 1 2\right)}$ and $\color{blue}{\left(0, -\dfrac 3 2, \dfrac 1 2\right)}$.
• $\boxed{\text{Sub case }y=0}$ It comes $0+0+3z~2-3=0$, that is, $z=\pm 1$, yielding the critical points $\color{blue}{(0,0,-1)}$ and $\color{blue}{(0,0,1)}$.

Now just check where it is hotter.