다변수 미분 전체 문제

(a) Since

$$\begin{aligned} f&=2x^3 - 6xy + y^2 +4y \\ f_x&=6x^2-6y & f_{xx}&=12x\qquad && f_{xy}&= -6\\ f_y&=-6x+2y+4\qquad & f_{yy}&=2 \end{aligned}$$

the critical points are the solutions of

$$\begin{aligned} & & &f_x=0\qquad & &f_y=0 \\ &\iff\qquad& &y=x^2\qquad & &y-3x+2=0 \\ &\iff& &y=x^2 & &x^2-3x+2=0 \\ &\iff& &y=x^2 & &x=1\text{ or }2 \end{aligned}$$

So, there are two critical points: $(1,1),\ (2,4)$.

$$\begin{array}{|c|c|c|c|}\hline \genfrac{}{}{0pt}{}{\text{critical}}{\text{point}} & f_{xx}f_{yy}-f_{xy}^2 & f_{xx} & \text{type} \\ \hline (1,1) & 12\times 2-(-6)^2<0 & & \text{saddle point} \\ \hline (2,4) & 24\times 2-(-6)^2>0 & 24 & \text{local min} \\ \hline \end{array}$$

(b) There are no critical points in the interior of the allowed region, so both the maximum and the minimum occur only on the boundary. The boundary consists of the line segments (i) $x=1$, $0\le y\le 1$, (ii) $y=1$, $0\le x\le 1$ and (iii) $y=1-x$, $0\le x\le 1$.

• First, we look at the part of the boundary with $x=1$. There $f=y^2-2y+2$. As $\frac{\mathrm{d}}{\mathrm{d}y}(y^2-2y+2)=2y-2$ vanishes only at $y=1$, the max and min of $y^2-2y+2$ for $0\le y\le 1$ must occur either at $y=0$, where $f=2$, or at $y=1$, where $f=1$.
• Next, we look at the part of the boundary with $y=1$. There $f=2x^3-6x+5$. As $\frac{\mathrm{d}}{\mathrm{d}x}(2x^3-6x+5)=6x^2-6$, the max and min of $2x^3-6x+5$ for $0\le x\le 1$ must occur either at $x=0$, where $f=5$, or at $x=1$, where $f=1$.
• Next, we look at the part of the boundary with $y=1-x$. There $f=2x^3-6x(1-x) +(1-x)^2+4(1-x)=2x^3+7x^2-12x+5$. As $\frac{\mathrm{d}}{\mathrm{d}x}(2x^3+7x^2-12x+5)=6x^2+14x-12 =2\big(3x^2+7x-6\big) =2(3x-2)(x+3)$, the max and min of $2x^3+7x^2-12x+5$ for $0\le x\le 1$ must occur either at $x=0$, where $f=5$, or at $x=1$, where $f=2$, or at $x=\frac{2}{3}$, where $f=2(\frac{8}{27})-6(\frac{2}{3})(\frac{1}{3})+\frac{1}{9}+\frac{4}{3} =\frac{16-36+3+36}{27} =\frac{19}{27}$.

So all together, we have the following candidates for max and min, with the max and min indicated.

$$\begin{array}{|c|c|c|c|c|}\hline \text{point} & (1,0) & (1,1) & (0,1) & \big(\frac{2}{3},\frac{1}{3}\big) \\ \hline \text{value of }f & 2 & 1 & 5 & \frac{19}{27} \\ \hline & & & \text{max} & \text{min} \\ \hline \end{array}$$

We have

$$\begin{aligned} f(x,y)&=x^4+y^4-4xy+2\quad & f_x(x,y)&=4x^3-4y\quad & f_{xx}(x,y)&=12x^2 \\ & & f_y(x,y)&=4y^3-4x & f_{yy}(x,y)&=12y^2 \\ & & & &f_{xy}(x,y)&=-4 \end{aligned}$$

At a critical point

$$\begin{aligned} f_x(x,y)=f_y(x,y)=0 &\iff y=x^3\text{ and }x=y^3 \\ &\iff x=x^9\text{ and }y=x^3 \\ &\iff x(x^8-1)=0,\ y=x^3 \\ &\iff (x,y)=(0,0)\text{ or }(1,1)\text{ or }(-1,-1) \end{aligned}$$

Here is a table giving the classification of each of the three critical points.

$$\begin{array}{|c|c|c|c|}\hline \genfrac{}{}{0pt}{}{\text{critical}}{\text{point}} & f_{xx}f_{yy}-f_{xy}^2 & f_{xx} & \text{type} \\ \hline (0,0) & 0\times 0-(-4)^2<0 & & \text{saddle point} \\ \hline (1,1) & 12\times 12-(-4)^2>0 & \text{12} & \text{local min} \\ \hline (-1,-1) & 12\times 12-(-4)^2>0 & \text{12} & \text{local min} \\ \hline \end{array}$$

(a) We have

$$\begin{aligned} f(x,y)&=xy(x+2y-6)\quad & f_x(x,y)&=2xy+2y^2-6y\quad & f_{xx}(x,y)&=2y \\ & & f_y(x,y)&=x^2+4xy-6x & f_{yy}(x,y)&=4x\\ & & & &f_{xy}(x,y)&=2x+4y-6 \end{aligned}$$

At a critical point

$$\begin{aligned} f_x(x,y)=f_y(x,y)=0 &\iff 2y(x+y-3)=0\text{ and }x(x+4y-6)=0 \\ &\iff \{y=0\text{ or }x+y=3\}\text{ and }\{x=0\text{ or }x+4y=6\}\\ &\iff \{x=y=0\}\text{ or }\{y=0,\ x+4y=6\}\\ &\text{ or }\{x+y=3,\ x=0\} \text{ or }\{x+y=3,\ x+4y=6\}\\ &\iff (x,y)=(0,0)\text{ or }(6,0)\text{ or }(0,3)\text{ or }(2,1) \end{aligned}$$

Here is a table giving the classification of each of the four critical points.

$$\begin{array}{|c|c|c|c|}\hline \genfrac{}{}{0pt}{}{\text{critical}}{\text{point}} & f_{xx}f_{yy}-f_{xy}^2 & f_{xx} & \text{type} \\ \hline (0,0) & 0\times 0-(-6)^2<0 & & \text{saddle point} \\ \hline (6,0) & 0\times 24-6^2<0 & & \text{saddle point} \\ \hline (0,3) & 6\times 0-6^2<0 & & \text{saddle point} \\ \hline (2,1) & 2\times 8-2^2>0 & \text{2} & \text{local min} \\ \hline \end{array}$$

(b) Observe that $xy=4$ and $x+2y=6$ intersect when $x=6-2y$ and

$$\begin{aligned} (6-2y)y=4 &\iff 2y^2-6y+4=0 \iff 2(y-1)(y-2)=0 \\ &\iff (x,y) = (4,1)\text{ or }(2,2) \end{aligned}$$

The shaded region in the sketch below is $D$.

None of the critical points are in $D$. So the max and min must occur at either $(2,2)$ or $(4,1)$ or on $xy=4$, $2<x<4$ (in which case $F(x)=f\big(x,\frac{4}{x}\big)=4\big(x+\frac{8}{x}-6)$ obeys $F'(x)=4-\frac{32}{x^2}=0\iff x=\pm2\sqrt{2}$) or on $x+2y=6$, $2<x<4$ (in which case $f(x,y)$ is identically zero). So the min and max must occur at one of

$$\begin{array}{|c|c|}\hline (x,y) & f(x,y) \\ \hline (2,2) & 2\times 2(2+2\times2-6)=0 \\ \hline (4,1) & 4\times 1(4+2\times 1-6)=0 \\ \hline (2\sqrt{2},\sqrt{2}) & 4(2\sqrt{2}+2\sqrt{2}-6)<0 \\ \hline \end{array}$$

The maximum value is $0$ and the minimum value is $4(4\sqrt{2}-6) \approx -1.37$.

We have

$$\begin{aligned} f(x,y)&=x^4+y^4-4xy\qquad & f_x(x,y)&=4x^3-4y\qquad & f_{xx}(x,y)&=12x^2\\ & & f_y(x,y)&=4y^3-4x & f_{yy}(x,y)&=12y^2 \\ & & & &f_{xy}(x,y)&=-4 \end{aligned}$$

At a critical point

$$\begin{aligned} f_x(x,y)=f_y(x,y)=0 &\iff y=x^3\text{ and }x=y^3 \iff x=x^9\text{ and }y=x^3 \\ &\iff x(x^8-1)=0,\ y=x^3\\ &\iff (x,y)=(0,0)\text{ or }(1,1)\text{ or }(-1,-1) \end{aligned}$$

Here is a table giving the classification of each of the three critical points.

$$\begin{array}{|c|c|c|c|}\hline \genfrac{}{}{0pt}{}{\text{critical}}{\text{point}} & f_{xx}f_{yy}-f_{xy}^2 & f_{xx} & \text{type} \\ \hline (0,0) & 0\times 0-(-4)^2<0 & & \text{saddle point} \\ \hline (1,1) & 12\times 12-(-4)^2>0 & \text{12} & \text{local min} \\ \hline (-1,-1) & 12\times 12-(-4)^2>0 & \text{12} & \text{local min} \\ \hline \end{array}$$

The coldest point must be either on the boundary of the plate or in the interior of the plate.

• On the semi–circular part of the boundary $0\le y\le 2$ and $x^2+y^2=4$ so that $T=\ln\big(1+x^2+y^2\big)-y=\ln 5-y$. The smallest value of $\ln 5-y$ is taken when $y$ is as large as possible, i.e. when $y=2$, and is $\ln 5 -2\approx -0.391$.
• On the flat part of the boundary, $y=0$ and $-2\le x\le 2$ so that $T=\ln\big(1+x^2+y^2\big)-y=\ln\big(1+x^2\big)$. The smallest value of $\ln\big(1+x^2\big)$ is taken when $x$ is as small as possible, i.e. when $x=0$, and is $0$.
• If the coldest point is in the interior of the plate, it must be at a critical point of $T(x,y)$. Since

$$\begin{aligned} T_x(x,y)=\frac{2x}{1+x^2+y^2}\qquad T_y(x,y)=\frac{2y}{1+x^2+y^2}-1 \end{aligned}$$

a critical point must have $x=0$ and $\frac{2y}{1+x^2+y^2}-1=0$, which is the case if and only if $x=0$ and $2y-1-y^2=0$. So the only critical point is $x=0,\ y=1$, where $T=\ln 2-1\approx -0.307$.

Since $-0.391<-0.307<0$, the coldest temperture is $-0.391$ and the coldest point is $(0,2)$.