What is the value of $k$?

Question

If the highest temperature $T$ at any point in space is $T(x,y,z)=kxyz^2$ on the surface of the sphere $x^2+y^2+z^2=a^2$ is $2a^4$ then the value of $k$ is...

(a)8

(b)12

(c)16

(d)32

I tried it by using Method of Lagrange Multiplier

I defined $L(x,y,z)=T+\lambda (x^2+y^2+z^2-a^2)$,then

$L_x=kyz^2+2x\lambda$

$L_y=kxz^2+2y\lambda$

$L_z=2kxyz^2+2z\lambda$

I got stuck here,please help in proceeding further

Thank you!

Answer 1

You seem to have forgotten the reason for finding $L_x$, $L_y$, and $L_z$! At a maximum (or minimum) they must be equal to 0! You have $L_x= kyz^2+ 2x\lambda= 0$, $L_y= kxz^2+ 2y\lambda= 0$, and $L_z= 2kxyz+ 2z\lambda= 0$. From that $kyz^2= -2x\lambda$, $kxz^2= -2y\lambda$, and $2kxyz= -2z\lambda$. Since a values of $\lambda$ is not necessary for the solution, I find it often simplest to eliminate $\lambda$ first by dividing one equation by another. The first equation divided by the second gives $\frac{kyz^2}{kxz^2}= \frac{-2x\lambda}{-2y\lambda}$ which reduces to $\frac{y}{z}= \frac{x}{y}$ or $y^2= xz$. The second equation divided by the third gives $\frac{kxz^2}{2kxyz}= \frac{-2y\lambda}{-2z\lambda}$ which reduces to $\frac{z}{2y}= \frac{y}{z}$ or $z^2= 2y^2$. Since $y^2= xz$, $z^2= 2xz$. Either z= 0 or z= x.

z= 0 gives y= 0 and then $x= \pm 1$. z= x gives $y= \pm z$ and the $x= y= z= \frac{a \sqrt{3}}{3}$. At that point T(x, y, z)= kxyz= k\frac{a^3\sqrt{3}}{9}= 2a^4$ so k= \frac{18a^4}{a^3\sqrt{3}}= 6a\sqrt{3}$.

Answer 2

For a short way use spherical coordinates

• the sphere is $r=a$
• temperature is $T=kr^4\sin \theta \cos \theta \sin^2 \phi \cos^2 \phi =kr^4\frac{\sin 2\theta}2\frac{\sin^22\phi}4\implies T_{max}=\frac{ka^4}8$

By lagrange multipliers note that

• $L_z=2kxyz+2z\lambda=0$ has solution for $z=0$ which leads to $x=y=z=0$

suppose $z\neq 0$ then

• $L_z=2kxyz+2z\lambda=0 \implies \lambda=-kxy$

and then

• $L_x=kyz^2+2x\lambda=0 \implies kyz^2-2kx^2y=0\implies z^2-2x^2=0$
• $L_y=kxz^2+2y\lambda=0 \implies kxz^2-2kxy^2=0 \implies z^2-2y^2=0$

that is

• $x^2=y^2 \implies x=\pm y \implies x^2+x^2+2x^2=a^2 \implies x=\pm \frac{a}2 \quad y=\pm \frac{a}2 \quad z=\pm \frac{a\sqrt2}2$ $\implies T_{max}=\frac{ka^4}8$

In both cases we have obtained $$T_{max}=\frac{ka^4}8=2a^4\implies k=16$$