Question

The temperature Tof the disk${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{⩽}\mathbf{1}$is given by $\mathit{T}\mathbf{=}\mathbf{2}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathit{y}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{x}$.Find the hottest and coldest points of the disk.

Short answer

The maximum value of T is 4 at$\left(1,0\right)$ and minimum$-\frac{16}{5}$ is at $\left(\frac{1}{5},\pm \frac{2}{5}\sqrt{6}\right)$.

Step-by-step solution

Given Information

The given function for temperature is$T=2x^2-3y^2-2x$ and the surface is $x^2+y^2⩽1$.

Important information

Lagrange multipliers can be used to find any problem's maximum and minimum.

Find the maximum z

Let$f=T=2x^2-3y^2-2x$ and also $\varphi \left(x,y\right)=x^2+y^2$.

Using the Lagrange multiplier$F\left(x,y\right)=2x^2-3y^2-2x+\lambda \left(x^2+y^2\right)$ is to be maximised.

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

$\frac{\partial F\left(x,y\right)}{\partial y}=-6y+\lambda \left(2y\right)$

$$\begin{gathered} -6y+\lambda \left(2y\right)=0 \\ -3y+\lambda y=0 \\ y=0 or \lambda =3 \end{gathered}$$

Differentiate$F\left(x,y\right)$ with respect to x and equate to 0 and solve when $\lambda =3$.

$\frac{\partial F\left(x,y\right)}{\partial x}=4x-2+\lambda \left(2x\right)$

$$\begin{gathered} 4x-2+\lambda \left(2x\right)=0 \\ 4x-2+3\left(2x\right)=0 \\ 10x=2 \\ x=\frac{1}{5} \end{gathered}$$

Substitute$\frac{1}{5}$ for x into$x^2+y^2=1$ and solve to find the value of y.

$$\begin{gathered} {\left(\frac{1}{5}\right)}^2+y^2=1 \\ {y}^2=\frac{24}{25} \\ y=\pm \frac{2}{5}\sqrt{6} \end{gathered}$$

Substitute 0 for y into $x^2+y^2=1$and solve to find the value of x.

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

The obtained points are$\left(\pm 1,0\right)$ and $\left(\frac{1}{5},\pm \frac{2}{5}\sqrt{6}\right)$.

Substitute the obtained values of x and y into $T=2x^2-3y^2-2x$.

$$\begin{gathered} T=2{\left(1\right)}^2-3{\left(0\right)}^2-2\left(1\right) \\ =0 \end{gathered}$$

And

$$\begin{gathered} T=2{\left(-1\right)}^2-3{\left(0\right)}^2-2\left(-1\right) \\ =4 \end{gathered}$$

or

$$\begin{gathered} T=2{\left(\frac{1}{5}\right)}^2-3{\left(\frac{2}{5}\sqrt{6}\right)}^2-2\left(\frac{1}{5}\right) \\ =\frac{2}{25}-\frac{72}{25}-\frac{2}{5} \\ =-\frac{16}{5} \end{gathered}$$

and

$$\begin{gathered} T=2{\left(\frac{1}{5}\right)}^2-3{\left(-\frac{2}{5}\sqrt{6}\right)}^2-2\left(\frac{1}{5}\right) \\ =\frac{2}{25}-\frac{72}{25}-\frac{2}{5} \\ =-\frac{16}{5} \end{gathered}$$

It can be observed that the maximum value of T is 4 at$\left(1,0\right)$ and minimum is$-\frac{16}{5}$ at $\left(\frac{1}{5},\pm \frac{2}{5}\sqrt{6}\right)$.