Answer to Question #275327 in Calculus for Hafsa Tahsin

Question

The temperature in space given by T(x, y, z)= 200xyz? Find the hottest temperature on a unit sphere centered at the origin. (DO NOT USE LAGRANGE MULTIPLIERS)

Answer 1

unit sphere centered at origin is $x^{2}+y^{2}+z^{2}=1$

 $\Rightarrow z^{2}=1-x^{2}-y^{2}$

given $T(x, y, z)=200 x y z^{2}$

$T(x, y)=200 x y\left(1-x^{2}-y^{2}\right)$

$\Rightarrow T(x, y)=200\left(x y-x^{3} y-x y^{3}\right)$

$T_{x}=200\left(y-3 x^{2} y-y^{3}\right), T_{y}=200\left(x-x^{3}-3 x y^{2}\right)$

for critical points $T_{x}=0, T_{y}=0$

 $\begin{aligned} &200\left(y-3 x^{2} y-y^{3}\right)=0,200\left(x-x^{3}-3 x y^{2}\right)=0 \\ &\Rightarrow 200 y\left(1-3 x^{2}-y^{2}\right)=0,200 x\left(1-x^{2}-3 y^{2}\right)=0 \\ &\Rightarrow y=0,1-3 x^{2}-y^{2}=0 ; x=0,1-x^{2}-3 y^{2}=0 \\ &y=0,1-x^{2}-3 y^{2}=0 \\ &\Rightarrow 1-x^{2}-3(0)^{2}=0 \\ &\Rightarrow x=-1,1 \end{aligned}$

 $\begin{aligned} &x=0,1-3 x^{2}-y^{2}=0 \\ &\Rightarrow 1-3\left(0^{2}\right)-y^{2}=0 \\ &\Rightarrow y=-1,1 \\ &1-3 x^{2}-y^{2}=0,1-x^{2}-3 y^{2}=0 \\ &\Rightarrow 3-9 x^{2}-3 y^{2}=0,1-x^{2}-3 y^{2}=0 \\ &\Rightarrow 3-9 x^{2}-3 y^{2}-1+x^{2}+3 y^{2}=0-0 \\ &\Rightarrow 2-8 x^{2}=0 \\ &\Rightarrow x^{2}=1 / 4 \\ &\Rightarrow x=-1 / 2,1 / 2 \end{aligned}$

 $\begin{aligned} &1-3 x^{2}-y^{2}=0, x^{2}=1 / 4 \\ &\Rightarrow 1-3(1 / 4)-y^{2}=0 \\ &\Rightarrow y^{2}=1 / 4 \\ &=y=-1 / 2,1 / 2 \\ &(-1,0),(1,0),(0,-1),(0,1),(-1 / 2,-1 / 2),(1 / 2,1 / 2),(-1 / 2,1 / 2),(1 / 2,-1 / 2)\\& \text { are only critical points on unit sphere } \\ &T(-1,0)=0, \\ &T(1,0)=0, \\ &T(0,-1)=0, \\ &T(0,1)=0, \\ &T(-1 / 2,-1 / 2)=200(-1 / 2)(-1 / 2)\left(1-(-1 / 2)^{2}-(-1 / 2)^{2}\right)=25, \\ &T(1 / 2,1 / 2)=200(1 / 2)(1 / 2)\left(1-(1 / 2)^{2}-(1 / 2)^{2}\right)=25 \\ &T(-1 / 2,1 / 2)=200(-1 / 2)(1 / 2)\left(1-(-1 / 2)^{2}-(1 / 2)^{2}\right)=-25, \\ &T(1 / 2,-1 / 2)=200(1 / 2)(-1 / 2)\left(1-(1 / 2)^{2}-(-1 / 2)^{2}\right)=-25 \\ &\text { hottest temperature on unit sphere is } 25 \text { units } \end{aligned}$