# Mathematics-I (3110014)

BE | Semester-1   Winter-2019 | 17-01-2020

## Q2) (b)

#### Discuss the Maxima and Minima of the function $3{\mathrm{x}}^{2}-{\mathrm{y}}^{2}+{\mathrm{x}}^{3}$.

$\mathrm{Here},\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=3{\mathrm{x}}^{2}-{\mathrm{y}}^{2}+{\mathrm{x}}^{3}.$

$\mathrm{Now},$

$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=\frac{\partial }{\partial \mathrm{x}}\left(3{\mathrm{x}}^{2}-{\mathrm{y}}^{2}+{\mathrm{x}}^{3}\right)$

$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=\frac{\partial }{\partial \mathrm{x}}\left(3{\mathrm{x}}^{2}\right)-\frac{\partial }{\partial \mathrm{x}}\left({\mathrm{y}}^{2}\right)+\frac{\partial }{\partial \mathrm{x}}\left({\mathrm{x}}^{3}\right)$

$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=6\mathrm{x}-0+3{\mathrm{x}}^{2}$

$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=6\mathrm{x}+3{\mathrm{x}}^{2}$

$\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=\frac{\partial }{\partial \mathrm{y}}\left(3{\mathrm{x}}^{2}-{\mathrm{y}}^{2}+{\mathrm{x}}^{3}\right)$

$\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=\frac{\partial }{\partial \mathrm{y}}\left(3{\mathrm{x}}^{2}\right)-\frac{\partial }{\partial \mathrm{y}}\left({\mathrm{y}}^{2}\right)+\frac{\partial }{\partial \mathrm{y}}\left({\mathrm{x}}^{3}\right)$

$\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=0-2y+0$

$\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=-2y$

$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=0$

$⟹6\mathrm{x}+3{\mathrm{x}}^{2}=0$

$⟹3\mathrm{x}\left(2+\mathrm{x}\right)=0$

$\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=0$

$⟹-2\mathrm{y}=0$

$⟹\mathrm{y}=0$

$\mathrm{Now},$

$\mathrm{r}=\frac{{\partial }^{2}\mathrm{f}}{\partial {\mathrm{x}}^{2}}$

${\mathrm{r}}=\frac{\partial }{\partial \mathrm{x}}\left(\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\right)$

${\mathrm{r}}=\frac{\partial }{\partial \mathrm{x}}\left(6\mathrm{x}+3{\mathrm{x}}^{2}\right)$

${\mathrm{r}}=\frac{\partial }{\partial \mathrm{x}}\left(6\mathrm{x}\right)+\frac{\partial }{\partial \mathrm{x}}\left(3{\mathrm{x}}^{2}\right)$

${\mathrm{r}}=6+6\mathrm{x}$

$\mathrm{s}=\frac{{\partial }^{2}\mathrm{f}}{\partial \mathrm{x}\partial \mathrm{y}}$

${\mathrm{s}}=\frac{\partial }{\partial \mathrm{x}}\left(\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\right)$

${\mathrm{s}}=\frac{\partial }{\partial \mathrm{x}}\left(-2\mathrm{y}\right)$

${\mathrm{s}}=0$

$\mathrm{t}=\frac{{\partial }^{2}\mathrm{f}}{\partial {\mathrm{y}}^{2}}$

${\mathrm{t}}=\frac{\partial }{\partial \mathrm{y}}\left(\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\right)$

${\mathrm{s}}=\frac{\partial }{\partial \mathrm{y}}\left(-2\mathrm{y}\right)$

${\mathrm{s}}=-2$

$\mathbf{Point}$ $\mathrm{r}=6+6\mathrm{x}$ $\mathbf{s}\mathbf{=}\mathbf{0}$ $\mathbf{r}\mathbf{=}\mathbf{-}\mathbf{2}$ $\mathbf{rt}\mathbf{-}{\mathbf{s}}^{\mathbf{2}}$ $\mathrm{\mathbf{Conclusion}}$
$\left(0,0\right)$ $6+6\left(0\right)=6>0$ $0$ $-2$ $6\left(-2\right)-\left(0{\right)}^{2}=-12<0$
$\left(-2,0\right)$ $6+6\left(-2\right)=-6<0$ $0$ $-2$ $-6\left(-2\right)-\left(0{\right)}^{2}=12>0$