91Ó°ÊÓ

The temperature at a point$\left(x,y,z\right)$ in the ball $x^2+y^2+z^2â\copyright ½1$is $T=y^2+xz$.

Lagrange multipliers can be utilized to find the maximum and minimum of any problem.

Let $f=T=y^2+xz$and also $\mathrm{Ï•}\left(x,y,z\right)=x^2+y^2+z^2$.

Using the Lagrange multiplier $F\left(x,y\right)=y^2+xz+\mathrm{λ}\left(x^2+y^2+z^2\right)$is to be maximised.

Differentiate $F\left(x,y,z\right)$with respect to y and equate to 0.

$\frac{∂F\left(x,y,z\right)}{∂y}=2y+\mathrm{λ}\left(2y\right)$

$$\begin{gathered} 2y+\mathrm{λ}\left(2y\right)=0 \\ y+\mathrm{λ}y=0 \\ y=0  or  \mathrm{λ}=-1 \end{gathered}$$

Differentiate $\mathrm{F}(x,y)$with respect to x and equate to 0.

$\frac{∂F\left(x,y\right)}{∂x}=z+\mathrm{λ}\left(2x\right)$

$$\begin{gathered} z+\mathrm{λ}\left(2x\right)=0 \\ \mathrm{λ}=-\frac{z}{2x} \end{gathered}$$

Differentiate $\mathrm{F}(x,y)$with respect to z and equate to 0.

$\frac{∂F\left(x,y\right)}{∂x}=x+\mathrm{λ}\left(2z\right)$

$$\begin{gathered} x+\mathrm{λ}\left(2z\right)=0 \\ \mathrm{λ}=-\frac{x}{2z} \end{gathered}$$

Equate all the obtained values of $\mathrm{λ}$and solve.

$$\begin{gathered} \mathrm{λ}=-\frac{x}{2z}=-\frac{z}{2x} \\ {x}^2=z^2 \end{gathered}$$

This implies that $x=Â\pm z$.

Substitute $z^2$for $x^2$into$x^2+z^2=1$ and solve to find the value of x.

$$\begin{gathered} x^2+x^2=1 \\ 2x^2=1 \\ x=Â\pm \sqrt{\frac{1}{2}} \end{gathered}$$

The obtained points are role="math" localid="1664877627937" $\left(\sqrt{\frac{1}{2}},\sqrt{\frac{1}{2}}\right)$, $\left(\sqrt{\frac{1}{2}},-\sqrt{\frac{1}{2}}\right)$,$\left(-\sqrt{\frac{1}{2}},\sqrt{\frac{1}{2}}\right)$ and $\left(-\sqrt{\frac{1}{2}},-\sqrt{\frac{1}{2}}\right)$.

Substitute the obtained values of x and z into $T=y^2+xz$.

$$\begin{gathered} T=0^2+\left(\sqrt{\frac{1}{2}}\right)\left(\sqrt{\frac{1}{2}}\right) \\ =\frac{1}{2} \end{gathered}$$

$$\begin{gathered} T=0^2+\left(-\sqrt{\frac{1}{2}}\right)\left(\sqrt{\frac{1}{2}}\right) \\ =-\frac{1}{2} \end{gathered}$$

$$\begin{gathered} T=0^2+\left(-\sqrt{\frac{1}{2}}\right)\left(-\sqrt{\frac{1}{2}}\right) \\ =\frac{1}{2} \end{gathered}$$

$$\begin{gathered} T=0^2+\left(\sqrt{\frac{1}{2}}\right)\left(-\sqrt{\frac{1}{2}}\right) \\ =-\frac{1}{2} \end{gathered}$$

It can be observed that the maximum value of T is$\frac{1}{2}$ and minimum is $-\frac{1}{2}$.

Equate the obtained value of the differentiation of $\mathrm{F}(x,y)$with respect to x to 0 and solve for $\mathrm{λ}=-1$.

$$\begin{gathered} z-\left(2x\right)=0 \\ z=2x \end{gathered}$$

Equate the obtained value of the differentiation of$\mathrm{F}(x,y)$ with respect to z to 0 and solve for $\mathrm{λ}=-1$.

$$\begin{gathered} x-1\left(2z\right)=0 \\ x=2z \end{gathered}$$

It can be observed that when $\mathrm{λ}=-1$, $x=z=0$. Substitute$x=z=0$ into$x^2+y^2+z^2=1$ to find the value of y.

$$\begin{gathered} 0^2+y^2+0^2=1 \\ {y}^2=1 \\ y=Â\pm 1 \end{gathered}$$

Substitute the obtained values of x, y and z into $T=y^2+xz$.

$$\begin{gathered} T={\left(Â\pm 1\right)}^2+\left(0\right)\left(0\right) \\ =1 \end{gathered}$$

It can be observed that the maximum value of T on surface is 1 and minimum is $-\frac{1}{2}$.

It can be observed that the maximum value of T on the whole ball is also 1 and minimum is $-\frac{1}{2}$.