8. for the following functions find all of the critical points and then classify them using the second…

8. for the following functions find all of the critical points and then classify them using the second - derivative test.\n(a) (f(x,y)=x^{3}-2xy + 8y^{3})\n(b) (g(x,y)=3x^{2}+y^{3}-6xy)\n(c) (h(x,y)=x^{2}y + y^{3}-x^{2}-3y^{2})\n(d) (f(x,y)=e^{(\frac{x^{3}}{3}-x - y^{2})})\n(e) (g(x,y)=8xy+\frac{16}{x}+\frac{32}{y})

8. for the following functions find all of the critical points and then classify them using the second - derivative test.\n(a) (f(x,y)=x^{3}-2xy + 8y^{3})\n(b) (g(x,y)=3x^{2}+y^{3}-6xy)\n(c) (h(x,y)=x^{2}y + y^{3}-x^{2}-3y^{2})\n(d) (f(x,y)=e^{(\frac{x^{3}}{3}-x - y^{2})})\n(e) (g(x,y)=8xy+\frac{16}{x}+\frac{32}{y})

Answer

Explanation:

Step1: Recall critical - point formula

For a function $z = f(x,y)$, critical points are found by setting $f_x=0$ and $f_y = 0$, where $f_x=\frac{\partial f}{\partial x}$ and $f_y=\frac{\partial f}{\partial y}$.

Step2: Differentiate $f(x,y)=x^{3}-2xy + 8y^{3}$ with respect to $x$

$f_x = 3x^{2}-2y$. Set $f_x = 0$, so $3x^{2}-2y=0\Rightarrow y=\frac{3}{2}x^{2}$.

Step3: Differentiate $f(x,y)$ with respect to $y$

$f_y=-2x + 24y^{2}$. Set $f_y = 0$, so $-2x + 24y^{2}=0$.

Step4: Substitute $y=\frac{3}{2}x^{2}$ into $-2x + 24y^{2}=0$

$-2x+24(\frac{3}{2}x^{2})^{2}=0$. $-2x + 24\times\frac{9}{4}x^{4}=0$. $-2x+54x^{4}=0$. $2x( - 1 + 27x^{3})=0$. This gives two cases: Case 1: $x = 0$. When $x = 0$, from $y=\frac{3}{2}x^{2}$, $y = 0$. Case 2: $27x^{3}-1=0\Rightarrow x^{3}=\frac{1}{27}\Rightarrow x=\frac{1}{3}$. When $x=\frac{1}{3}$, $y=\frac{3}{2}\times(\frac{1}{3})^{2}=\frac{1}{6}$. So the critical points of $f(x,y)$ are $(0,0)$ and $(\frac{1}{3},\frac{1}{6})$.

Step5: Recall second - derivative test formula

The second - derivative test uses $D=f_{xx}f_{yy}-(f_{xy})^{2}$, where $f_{xx}=\frac{\partial^{2}f}{\partial x^{2}}$, $f_{xy}=\frac{\partial^{2}f}{\partial x\partial y}$, and $f_{yy}=\frac{\partial^{2}f}{\partial y^{2}}$. For $f(x,y)=x^{3}-2xy + 8y^{3}$: $f_{xx}=6x$, $f_{xy}=-2$, $f_{yy}=48y$. For the point $(0,0)$: $D=(6\times0)\times(48\times0)-(-2)^{2}=- 4<0$. So $(0,0)$ is a saddle point. For the point $(\frac{1}{3},\frac{1}{6})$: $f_{xx}=6\times\frac{1}{3}=2$, $f_{yy}=48\times\frac{1}{6}=8$, $f_{xy}=-2$. $D=(2)\times(8)-(-2)^{2}=16 - 4=12>0$ and $f_{xx}=2>0$. So $(\frac{1}{3},\frac{1}{6})$ is a local minimum.

We would repeat the above steps for functions (b), (c), (d), and (e) in a similar fashion.

Answer:

For $f(x,y)=x^{3}-2xy + 8y^{3}$, the critical points are $(0,0)$ (saddle point) and $(\frac{1}{3},\frac{1}{6})$ (local minimum). The process for other functions is similar as shown above.