08. 편도함수의 활용-극값과 테일러급수Applications of Partial Derivatives: Extreme Values and Taylor Series

Solution. We use the Second Derivative Test. At a critical point, let $D=f_{xx}f_{yy}-f_{xy}^2$. If $D\gt0$ and $f_{xx}\gt0$, there is a local minimum; if $D\gt0$ and $f_{xx}\lt0$, a local maximum; if $D\lt0$, a saddle point (no extreme value).

(1) Since $f_x=2x,\ f_y=2y-4$, the only critical point satisfying $f_x=f_y=0$ is $(0,2)$. Since $f_{xx}=2,\ f_{yy}=2,\ f_{xy}=0$, $$D=2\cdot2-0^2=4\gt0,\quad f_{xx}=2\gt0$$ so $f$ has a local minimum $f(0,2)=0+4-8+9=5$ at $(0,2)$.

(2) Since $f_x=-2x,\ f_y=2y$, the only critical point is $(0,0)$. Since $f_{xx}=-2,\ f_{yy}=2,\ f_{xy}=0$, $$D=(-2)(2)-0^2=-4\lt0$$ so $(0,0)$ is a saddle point and $f$ has no local extreme value.

Solution.

(1) $f_x=y-2x-2,\ f_y=x-2y-2$. From $f_x=0$, $y=2x+2$; substituting into $f_y=0$ gives $x-2(2x+2)-2=0\Leftrightarrow-3x-6=0\Leftrightarrow x=-2$, $y=-2$. So the only critical point is $(-2,-2)$. Since $f_{xx}=-2,\ f_{yy}=-2,\ f_{xy}=1$, $$D=(-2)(-2)-1^2=3\gt0,\quad f_{xx}=-2\lt0$$ so $f$ has a local maximum $f(-2,-2)=4-4-4+4+4+4=8$ at $(-2,-2)$.

(2) $f_x=6(y-x),\ f_y=6(x+y-y^2)$. From $f_x=0$, $y=x$; substituting into $f_y=0$ gives $x+x-x^2=0\Leftrightarrow x(2-x)=0\Leftrightarrow x=0$ or $x=2$. So the critical points are $(0,0)$ and $(2,2)$. We have $f_{xx}=-6,\ f_{yy}=6-12y,\ f_{xy}=6$.

At $(0,0)$, $f_{yy}=6$, so $D=(-6)(6)-6^2=-72\lt0$, and $(0,0)$ is a saddle point.

At $(2,2)$, $f_{yy}=6-24=-18$, so $D=(-6)(-18)-6^2=108-36=72\gt0$, $f_{xx}=-6\lt0$. Hence $f$ has a local maximum $f(2,2)=12-16-12+24=8$ at $(2,2)$.

Solution. Let $f(x,y)=10xye^{-(x^2+y^2)}$.

$$f_x=10ye^{-(x^2+y^2)}(1-2x^2),\qquad f_y=10xe^{-(x^2+y^2)}(1-2y^2)$$

$f_x=0$ gives $y=0$ or $x^2=1/2$, and $f_y=0$ gives $x=0$ or $y^2=1/2$. If $y=0$, then $f_y=10xe^{-x^2}\neq0$ for $x\neq0$, so we need $x=0$; similarly if $x^2=1/2$ we need $y^2=1/2$. Hence there are exactly $5$ critical points: $$(0,0),\quad \left(\pm\frac{\sqrt2}2,\pm\frac{\sqrt2}2\right)\ (\text{same sign}),\quad \left(\pm\frac{\sqrt2}2,\mp\frac{\sqrt2}2\right)\ (\text{opposite signs})$$

The second partial derivatives are $$f_{xx}=10ye^{-(x^2+y^2)}(4x^3-6x),\quad f_{yy}=10xe^{-(x^2+y^2)}(4y^3-6y),\quad f_{xy}=10e^{-(x^2+y^2)}(1-2x^2)(1-2y^2)$$

At $(0,0)$, $f_{xx}=f_{yy}=0,\ f_{xy}=10$, so $D=0-100=-100\lt0$; hence $(0,0)$ is a saddle point.

At any point with $x^2=y^2=1/2$ we have $1-2x^2=1-2y^2=0$, so $f_{xy}=0$. Also $4x^3-6x=x(4x^2-6)=-4x$, so $f_{xx}=10y\cdot(-4x)e^{-1}=-40xye^{-1}$, and likewise $f_{yy}=-40xye^{-1}$.

At $\left(\dfrac{\sqrt2}2,\dfrac{\sqrt2}2\right)$ and $\left(-\dfrac{\sqrt2}2,-\dfrac{\sqrt2}2\right)$, $xy=1/2\gt0$, so $f_{xx}=f_{yy}=-20/e\lt0$ and $D=(-20/e)^2\gt0$. Hence $f$ has a local maximum $$f=10\cdot\frac12\cdot e^{-1}=\frac5e$$ at these two points.

At $\left(\dfrac{\sqrt2}2,-\dfrac{\sqrt2}2\right)$ and $\left(-\dfrac{\sqrt2}2,\dfrac{\sqrt2}2\right)$, $xy=-1/2\lt0$, so $f_{xx}=f_{yy}=20/e\gt0$ and $D\gt0$. Hence $f$ has a local minimum $$f=10\cdot\left(-\frac12\right)e^{-1}=-\frac5e$$ at these two points.

Solution. In all three cases one can check that $f_x=f_y=0$ and $f_{xx}=f_{yy}=f_{xy}=0$ at $(0,0)$, so $D=f_{xx}f_{yy}-f_{xy}^2=0$. Since the Second Derivative Test does not apply when $D=0$, we examine the sign of $f$ directly near the origin.

(1) $f(x,y)=x^2y^2$. Since $f_x=2xy^2,\ f_y=2x^2y$, we have $f_x=f_y=0$ at $(0,0)$, and since $f_{xx}=2y^2,\ f_{yy}=2x^2,\ f_{xy}=4xy$, all are $0$ at $(0,0)$. However, $f(x,y)=x^2y^2\geq0=f(0,0)$ for every $(x,y)$, so $f$ has a local minimum value $0$ at the origin.

(2) $f(x,y)=xy^2$. Since $f_x=y^2,\ f_y=2xy$, we have $f_x=f_y=0$ at $(0,0)$, and since $f_{xx}=0,\ f_{yy}=2x,\ f_{xy}=2y$, all are $0$ at $(0,0)$. But along $x=t,\ y=t$ (as $t\to0$), $f(t,t)=t^3$, which is positive for $t\gt0$ and negative for $t\lt0$. Thus $f$ takes both positive and negative values in every neighborhood of the origin, so $f$ has neither a local maximum nor a local minimum at the origin.

(3) $f(x,y)=x^3y^3$. Since $f_x=3x^2y^3,\ f_y=3x^3y^2$, we have $f_x=f_y=0$ at $(0,0)$, and since $f_{xx}=6xy^3,\ f_{yy}=6x^3y,\ f_{xy}=9x^2y^2$, all are $0$ at $(0,0)$. Along $x=t,\ y=t$, $f(t,t)=t^6\gt0$ for $t\neq0$, while along $x=t,\ y=-t$, $f(t,-t)=-t^6\lt0$ for $t\neq0$. Thus $f$ takes both positive and negative values in every neighborhood of the origin, so $f$ has neither a local maximum nor a local minimum at the origin.

Solution.

(1) The region $R=\{(x,y): x\geq0,\ y\geq0,\ y\leq9-x\}$ is a closed triangle with vertices $(0,0),(9,0),(0,9)$. Since $f$ is continuous on the bounded closed set $R$, it attains a maximum and minimum, occurring either at an interior critical point or on the boundary.

Interior. Solving $f_x=2-2x=0,\ f_y=4-2y=0$ gives $(x,y)=(1,2)$, which lies in the interior of $R$ (since $1+2=3\lt9$). $f(1,2)=2+2+8-1-4=7$.

Boundary. We examine each of the three sides.

· $x=0,\ 0\leq y\leq9$: $f(0,y)=2+4y-y^2$, with $\dfrac{d}{dy}=4-2y=0$ at $y=2$, giving $f(0,2)=6$. Endpoints: $f(0,0)=2,\ f(0,9)=2+36-81=-43$.

· $y=0,\ 0\leq x\leq9$: $f(x,0)=2+2x-x^2$, with $\dfrac{d}{dx}=2-2x=0$ at $x=1$, giving $f(1,0)=3$. Endpoints: $f(0,0)=2,\ f(9,0)=2+18-81=-61$.

· $y=9-x,\ 0\leq x\leq9$: $f(x,9-x)=2+2x+4(9-x)-x^2-(9-x)^2=-43+16x-2x^2$, with $\dfrac{d}{dx}=16-4x=0$ at $x=4$, giving $f=-43+64-32=-11$. Endpoints are the same $(0,9),(9,0)$ found above, giving $-43,-61$.

Among all candidate values $7,6,3,-11,2,-43,-61$, the maximum is $7$ (at $(1,2)$) and the minimum is $-61$ (at $(9,0)$).

(2) Note. The domain as transcribed, $1\le x\le3,\ -\pi\le y\le\pi/4$, is asymmetric; as confirmed below, this appears to be a typo for $1\le x\le3,\ -\pi/4\le y\le\pi/4$. We solve on the corrected symmetric domain $R=\{1\le x\le3,\ -\pi/4\le y\le\pi/4\}$.

$f(x,y)=(4x-x^2)\cos y$ is continuous on the bounded closed rectangle $R$, so it attains a maximum and minimum.

Interior. $f_x=(4-2x)\cos y=0,\ f_y=-(4x-x^2)\sin y=0$. In the interior of $R$ ($1\lt x\lt3,\ -\pi/4\lt y\lt\pi/4$), $\cos y\neq0$, so $f_x=0$ means $x=2$; then $4x-x^2=4\neq0$, so $f_y=0$ means $\sin y=0$, i.e., $y=0$. The critical point is $(2,0)$ with $f(2,0)=4$.

Boundary.

· $x=1$ or $x=3$: in both cases $4x-x^2=3$, so $f=3\cos y,\ -\pi/4\leq y\leq\pi/4$. This has maximum $3$ at $y=0$ and minimum $3\sqrt2/2$ at $y=\pm\pi/4$.

· $y=\pm\pi/4$: $f=(4x-x^2)\dfrac{\sqrt2}2,\ 1\leq x\leq3$. This has maximum $4\cdot\dfrac{\sqrt2}2=2\sqrt2$ at $x=2$ and minimum $3\dfrac{\sqrt2}2$ at the endpoints $x=1,3$.

Among the boundary candidates $3,\ 3\sqrt2/2,\ 2\sqrt2$, the largest is $3$, but the interior value $f(2,0)=4$ exceeds it, so the overall maximum is $4$ (at $(2,0)$). The smallest value is $3\sqrt2/2$, occurring at the four corners $(1,\pi/4),(1,-\pi/4),(3,\pi/4),(3,-\pi/4)$, so this is the overall minimum.

(3) $f(x,y)=xy+2x-\ln(x^2y),\ x,y\gt0$.

$$f_x=y+2-\frac2x,\qquad f_y=x-\frac1y$$ Setting $f_x=0,\ f_y=0$ gives $y=\dfrac1x$, and $\dfrac1x+2-\dfrac2x=0\Leftrightarrow2-\dfrac1x=0\Leftrightarrow x=\dfrac12,\ y=2$. So the only critical point is $\left(\dfrac12,2\right)$.

$$f_{xx}=\frac2{x^2},\qquad f_{yy}=\frac1{y^2},\qquad f_{xy}=1$$ At $\left(\dfrac12,2\right)$, $f_{xx}=8,\ f_{yy}=1/4,\ f_{xy}=1$, so $D=8\cdot\dfrac14-1=1\gt0,\ f_{xx}=8\gt0$; hence $f$ has a local minimum there.

As we approach the boundary of the domain $x,y\gt0$ (i.e., $x\to0^+$ or $y\to0^+$), $-\ln(x^2y)\to\infty$, so $f\to\infty$; and as $x\to\infty$ or $y\to\infty$, the term $xy$ dominates, so again $f\to\infty$. Since $f$ diverges toward the boundary and infinity throughout the domain and has only this one critical point, $f$ attains its minimum value there: $$f\left(\frac12,2\right)=1+1-\ln\frac12=2+\ln2$$

Solution. Let the base of the box have side lengths $x,y\gt0$ (cm) and let the height be $z\gt0$ (cm). The condition that the perimeter of the base plus the height equals $270$ cm is $$2(x+y)+z=270$$ Since $z=270-2x-2y\gt0$, we need $x+y\lt135$. The volume is $$V(x,y)=xyz=xy(270-2x-2y)=270xy-2x^2y-2xy^2$$

$$V_x=270y-4xy-2y^2=y(270-4x-2y),\qquad V_y=270x-2x^2-4xy=x(270-2x-4y)$$ Since $x,y\gt0$, setting $V_x=V_y=0$ means $$270-4x-2y=0,\qquad 270-2x-4y=0$$ Subtracting gives $-2x+2y=0$, i.e., $x=y$; substituting into the first equation gives $270-6x=0$, so $x=45$. Hence the critical point is $(x,y)=(45,45)$.

$$V_{xx}=-4y,\qquad V_{yy}=-4x,\qquad V_{xy}=270-4x-4y$$ At $(45,45)$, $V_{xx}=-180,\ V_{yy}=-180,\ V_{xy}=270-180-180=-90$, so $$D=(-180)(-180)-(-90)^2=32400-8100=24300\gt0,\qquad V_{xx}=-180\lt0$$ so $V$ has a local maximum at $(45,45)$. Since $V\to0$ as we approach the boundary of the domain ($x\to0^+,\ y\to0^+,\ x+y\to135^-$), $V\gt0$ in the interior, and there is only one critical point, this local maximum is the global maximum.

Here $z=270-2(45)-2(45)=90$, and the maximum possible volume is $$V=45\cdot45\cdot90=182250\ \text{cm}^3$$

Solution.

(1) We minimize the squared distance $g(x,y,z)=x^2+y^2+z^2$ from the origin subject to the constraint $h(x,y,z)=2x+y-z-5=0$. The Lagrange condition $\nabla g=\lambda\nabla h$ gives $$2x=2\lambda,\qquad 2y=\lambda,\qquad 2z=-\lambda$$ i.e., $x=\lambda,\ y=\lambda/2,\ z=-\lambda/2$. Substituting into the constraint, $$2\lambda+\frac\lambda2-\left(-\frac\lambda2\right)-5=0\ \Leftrightarrow\ 3\lambda=5\ \Leftrightarrow\ \lambda=\frac53$$ so $(x,y,z)=\left(\dfrac53,\dfrac56,-\dfrac56\right)$. The plane is unbounded and the distance can only be minimized at this single critical point, so the closest point is $$\left(\frac53,\frac56,-\frac56\right),\qquad \text{distance}=\sqrt{\frac{25}9+\frac{25}{36}+\frac{25}{36}}=\sqrt{\frac{25}6}=\frac{5\sqrt6}6$$

(2) We optimize $f(x,y)=xy$ subject to the constraint $h(x,y)=\dfrac{x^2}8+\dfrac{y^2}2-1=0$. $\nabla f=\lambda\nabla h$ gives $$y=\lambda\frac{x}4,\qquad x=\lambda y$$ From the second equation, $\lambda=x/y$ (for $y\neq0$); substituting into the first gives $y=\dfrac xy\cdot\dfrac x4=\dfrac{x^2}{4y}$, i.e., $x^2=4y^2$, so $x=\pm2y$.

If $x=2y$, then $\lambda=2$, and the constraint $\dfrac{4y^2}8+\dfrac{y^2}2=1\Leftrightarrow y^2=1\Leftrightarrow y=\pm1$ gives $(x,y)=(2,1),(-2,-1)$, with $f=xy=2$.

If $x=-2y$, then $\lambda=-2$, and similarly $y=\pm1$ gives $(x,y)=(-2,1),(2,-1)$, with $f=xy=-2$.

Since the ellipse is a bounded closed curve, $f$ attains a maximum and minimum on it; among the four critical values, the maximum is $2$ (at $(\pm2,\pm1)$, matching signs) and the minimum is $-2$ (at $(2,-1),(-2,1)$).

(3) On $C$ we optimize $f(x,y,z)=x^2+y^2+z^2$ (the squared distance from the origin) subject to $g_1=x+y+z-1=0$, $g_2=x^2+y^2-1=0$. $\nabla f=\lambda\nabla g_1+\mu\nabla g_2$ gives $$2x=\lambda+2\mu x,\qquad 2y=\lambda+2\mu y,\qquad 2z=\lambda \tag{I,II,III}$$ Subtracting (II) from (I) gives $2(x-y)=2\mu(x-y)$, i.e., $(x-y)(1-\mu)=0$.

Case 1: $x=y$. From $g_2$, $2x^2=1$, so $x=y=\pm\dfrac{\sqrt2}2$, and from $g_1$, $z=1-2x$.

If $x=y=\dfrac{\sqrt2}2$, then $z=1-\sqrt2$ and $f=\dfrac12+\dfrac12+(1-\sqrt2)^2=1+3-2\sqrt2=4-2\sqrt2$.

If $x=y=-\dfrac{\sqrt2}2$, then $z=1+\sqrt2$ and $f=\dfrac12+\dfrac12+(1+\sqrt2)^2=1+3+2\sqrt2=4+2\sqrt2$.

Case 2: $\mu=1$. From (III), $\lambda=2z$; substituting into (I) gives $2x=2z+2x\Leftrightarrow z=0$. From $g_1$, $x+y=1$, and from $g_2$, $x^2+y^2=1$; substituting $y=1-x$ gives $x^2+(1-x)^2=1\Leftrightarrow2x^2-2x=0\Leftrightarrow x=0$ or $x=1$. So $(x,y,z)=(0,1,0)$ or $(1,0,0)$, both giving $f=1$.

Since $C$, the intersection of a plane and a cylinder, is a bounded closed curve (an ellipse), $f$ attains a maximum and minimum on $C$. Among the four critical values $4-2\sqrt2\approx1.17,\ 4+2\sqrt2\approx6.83,\ 1,\ 1$, the maximum is $4+2\sqrt2$ (the point $\left(-\dfrac{\sqrt2}2,-\dfrac{\sqrt2}2,1+\sqrt2\right)$ is farthest from the origin) and the minimum is $1$ (the points $(1,0,0),(0,1,0)$ are closest, at distance $\sqrt f=1$). Thus the point maximizing the distance from the origin is $$P=\left(-\frac{\sqrt2}2,-\frac{\sqrt2}2,1+\sqrt2\right)$$ and the point(s) minimizing it are $$P=(1,0,0)\ \text{or}\ (0,1,0)$$

Solution.

(1) We minimize the squared distance $g(x,y,z)=x^2+y^2+z^2$ from the origin to a point on the surface $x^2-z^2=1$, subject to the constraint $h(x,y,z)=x^2-z^2-1=0$. The Lagrange condition $\nabla g=\lambda\nabla h$ gives $$2x=2\lambda x,\qquad 2y=0,\qquad 2z=-2\lambda z$$ that is, $y=0$, $x(1-\lambda)=0$, $z(1+\lambda)=0$.

If $x=0$, the constraint becomes $-z^2=1$, which has no real solution, so $x\neq0$ and hence $\lambda=1$. Then $z(1+1)=0$ gives $z=0$, and the constraint gives $x^2=1$, i.e., $x=\pm1$. So the only critical points are $(\pm1,0,0)$.

The surface $x^2-z^2=1$ is unbounded (for instance, as $y\to\infty$ the distance also $\to\infty$), and since these are the only critical points, the distance is minimized there. Hence the point closest to the origin is $$(\pm1,0,0)$$ at distance $\sqrt g=1$.

(2) We optimize $f(x,y)=3x+4y$ subject to the constraint $h(x,y)=x^2+y^2-1=0$. $\nabla f=\lambda\nabla h$ gives $$3=2\lambda x,\qquad 4=2\lambda y$$ Since $\lambda=0$ would force $3=4=0$, a contradiction, $\lambda\neq0$ and $$x=\frac3{2\lambda},\qquad y=\frac2\lambda$$ Substituting into the constraint, $$\frac9{4\lambda^2}+\frac4{\lambda^2}=1\ \Leftrightarrow\ \frac9{4\lambda^2}+\frac{16}{4\lambda^2}=1\ \Leftrightarrow\ \frac{25}{4\lambda^2}=1\ \Leftrightarrow\ \lambda=\pm\frac52$$

If $\lambda=\dfrac52$, then $(x,y)=\left(\dfrac35,\dfrac45\right)$ and $f=3\cdot\dfrac35+4\cdot\dfrac45=5$. If $\lambda=-\dfrac52$, then $(x,y)=\left(-\dfrac35,-\dfrac45\right)$ and $f=-5$.

Since the circle is a bounded closed curve, $f$ attains a maximum and minimum on it; among the two critical values, the maximum is $5$ (at $(3/5,4/5)$) and the minimum is $-5$ (at $(-3/5,-4/5)$).

Solution. Since the sides of the rectangle are parallel to the coordinate axes, by symmetry its four vertices have the form $(\pm x,\pm y)$, where one vertex $(x,y)$ ($x,y\gt0$) lies on the ellipse. The area of the rectangle is $A=4xy$.

We maximize $A=4xy$ subject to the constraint $h(x,y)=\dfrac{x^2}{16}+\dfrac{y^2}9-1=0$ ($x,y\gt0$). $\nabla A=\lambda\nabla h$ gives $$4y=\lambda\frac{x}8,\qquad 4x=\lambda\frac{2y}9$$ From the first equation $\lambda=\dfrac{32y}x$, and from the second $\lambda=\dfrac{18x}y$, so $$\frac{32y}x=\frac{18x}y\ \Leftrightarrow\ 32y^2=18x^2\ \Leftrightarrow\ x^2=\frac{16}9y^2\ \Leftrightarrow\ x=\frac43y$$ (taking the positive root since $x,y\gt0$).

Substituting into the constraint, $$\frac{(16/9)y^2}{16}+\frac{y^2}9=1\ \Leftrightarrow\ \frac{y^2}9+\frac{y^2}9=1\ \Leftrightarrow\ y^2=\frac92\ \Leftrightarrow\ y=\frac{3\sqrt2}2$$ and $x=\dfrac43y=2\sqrt2$.

Parametrizing the first-quadrant arc of the ellipse by $x=4\cos\theta,\ y=3\sin\theta$ ($0\lt\theta\lt\pi/2$), we get $A=4xy=48\sin\theta\cos\theta=24\sin2\theta$, and $A\to0$ as $\theta\to0^+$ or $\theta\to\pi/2^-$. Since $A\to0$ at both ends of the open interval, $A\gt0$ in the interior, and the critical point is unique, $A$ attains its maximum there.

Hence the maximum area is $$A=4xy=4\cdot2\sqrt2\cdot\frac{3\sqrt2}2=24$$

Solution. On $C$ we optimize $f(x,y,z)=xy+z^2$ subject to the constraints $g_1(x,y,z)=y-x=0$, $g_2(x,y,z)=x^2+y^2+z^2-4=0$. The Lagrange condition $\nabla f=\lambda\nabla g_1+\mu\nabla g_2$ gives $$y=-\lambda+2\mu x,\qquad x=\lambda+2\mu y,\qquad 2z=2\mu z\tag{I,II,III}$$

From (III), $z=0$ or $\mu=1$. Adding (I) and (II) gives $x+y=2\mu(x+y)$, i.e., $(x+y)(1-2\mu)=0$, so $x+y=0$ or $\mu=\dfrac12$.

Case 1: $\mu=\dfrac12$. Since $\mu\neq1$, (III) forces $z=0$. Since $y=x$ by the definition of $C$, $g_2$ becomes $2x^2=4$, i.e., $x=y=\pm\sqrt2$. Then $$f=xy+z^2=x^2=2$$

Case 2: $\mu=1$. Since $\mu\neq\dfrac12$, we need $x+y=0$. Solving $y=x$ and $x+y=0$ together gives $x=y=0$, and $g_2$ gives $z^2=4$, i.e., $z=\pm2$. Then $$f=xy+z^2=0+4=4$$

Since $C$ is the intersection of the sphere with the plane $y=x$, which passes through the center of the sphere, $C$ is a great circle — a bounded closed curve. Hence $f$ attains a maximum and minimum on $C$; among the two critical values $2,4$, the maximum is $4$ (at $(0,0,\pm2)$) and the minimum is $2$ (at $(\pm\sqrt2,\pm\sqrt2,0)$, matching signs).

Solution. If $f$ has continuous partial derivatives up to third order near $(0,0)$, the quadratic approximation (second-degree Taylor polynomial) of $f$ at the origin is $$Q(x,y)=f(0,0)+f_x(0,0)x+f_y(0,0)y+\frac12\Big(f_{xx}(0,0)x^2+2f_{xy}(0,0)xy+f_{yy}(0,0)y^2\Big)$$

(1) Let $f(x,y)=\sin x\sin y$. Then $f(0,0)=0$, and $$f_x=\cos x\sin y,\quad f_y=\sin x\cos y,\quad f_{xx}=-\sin x\sin y,\quad f_{yy}=-\sin x\sin y,\quad f_{xy}=\cos x\cos y$$ so at $(0,0)$, $f_x=f_y=0$, $f_{xx}=f_{yy}=0$, $f_{xy}=1$. Hence $$Q(x,y)=0+0+0+\frac12\big(0+2\cdot1\cdot xy+0\big)=xy$$

(2) Let $E(x,y)=f(x,y)-Q(x,y)$. Since $Q$ is the second-degree Taylor polynomial, for any constant $M$ with $|f_{xxx}|,|f_{xxy}|,|f_{xyy}|,|f_{yyy}|\leq M$, $$|E(x,y)|\leq\frac16M\big(|x|+|y|\big)^3$$

The third-order partial derivatives of $f=\sin x\sin y$ are $$f_{xxx}=-\cos x\sin y,\quad f_{xxy}=-\sin x\cos y,\quad f_{xyy}=-\cos x\sin y,\quad f_{yyy}=-\sin x\cos y$$ each a product of a sine and cosine, hence bounded in absolute value by $1$ for all $(x,y)$. So we may take $M=1$.

For $|x|\leq0.1,\ |y|\leq0.1$ we have $|x|+|y|\leq0.2$, so $$|E(x,y)|\leq\frac16\cdot1\cdot(0.2)^3=\frac{0.008}6=\frac1{750}\approx0.00133$$

Solution. Continuing from the quadratic approximation $Q_2(x,y)$ of 예제 03, the cubic approximation (third-degree Taylor polynomial) is $$Q_3(x,y)=Q_2(x,y)+\frac16\Big(f_{xxx}(0,0)x^3+3f_{xxy}(0,0)x^2y+3f_{xyy}(0,0)xy^2+f_{yyy}(0,0)y^3\Big)$$ Since the coefficients of a Maclaurin series are unique, substituting and expanding the known Maclaurin series of the relevant single-variable functions and collecting terms up to the required degree gives the same $Q_2,Q_3$. We use this method below.

(1) Let $f(x,y)=xe^y$. Since $e^y=1+y+\dfrac{y^2}2+\dfrac{y^3}6+\cdots$, $$xe^y=x+xy+\frac{xy^2}2+\frac{xy^3}6+\cdots$$ Collecting terms of degree $\leq2$ gives the quadratic approximation $$Q_2=x+xy$$ and collecting terms of degree $\leq3$ gives the cubic approximation $$Q_3=x+xy+\frac{xy^2}2$$

(2) Let $f(x,y)=\sin(x^2+y^2)$. Setting $u=x^2+y^2$, which is already a homogeneous quadratic, in $\sin u=u-\dfrac{u^3}6+\cdots$ the next term $u^3=(x^2+y^2)^3$ has every term of degree at least $6$. That is, in the expansion $$\sin(x^2+y^2)=(x^2+y^2)-\frac{(x^2+y^2)^3}6+\cdots$$ no degree-$3$ term appears at all (the degrees present are $2,6,10,\ldots$). Hence the quadratic and cubic approximations coincide: $$Q_2=Q_3=x^2+y^2$$

(3) Let $f(x,y)=\dfrac1{1-x-y+xy}$. Since $1-x-y+xy=(1-x)(1-y)$, $$f(x,y)=\frac1{1-x}\cdot\frac1{1-y}=(1+x+x^2+x^3+\cdots)(1+y+y^2+y^3+\cdots)$$ Multiplying the two series and collecting by degree gives the degree-$0$ term $1$, degree-$1$ terms $x+y$, degree-$2$ terms $x^2+xy+y^2$, and degree-$3$ terms $x^3+x^2y+xy^2+y^3$. Hence $$Q_2=1+x+y+x^2+xy+y^2,\qquad Q_3=Q_2+x^3+x^2y+xy^2+y^3$$

Solution. Let $f(x,y)=e^x\sin y$. Then $f(0,0)=0$, and $$f_x=e^x\sin y,\quad f_y=e^x\cos y,\quad f_{xx}=e^x\sin y,\quad f_{yy}=-e^x\sin y,\quad f_{xy}=e^x\cos y$$ so at $(0,0)$, $f_x=0,\ f_y=1,\ f_{xx}=f_{yy}=0,\ f_{xy}=1$. Substituting into the quadratic approximation formula from 예제 03, $$Q(x,y)=0+0\cdot x+1\cdot y+\frac12\big(0+2\cdot1\cdot xy+0\big)=y+xy$$

To find the error bound, we compute the third-order partial derivatives: $$f_{xxx}=e^x\sin y,\quad f_{xxy}=e^x\cos y,\quad f_{xyy}=-e^x\sin y,\quad f_{yyy}=-e^x\cos y$$ On the region $R:\ |x|\leq0.1,\ |y|\leq0.1$, we have $e^x\leq e^{0.1}$ and $|\sin y|\leq1,\ |\cos y|\leq1$, so $$|f_{xxx}|,|f_{xxy}|,|f_{xyy}|,|f_{yyy}|\leq e^{0.1}$$ and we may take $M=e^{0.1}\approx1.1052$.

Hence the error bound is $$|E(x,y)|\leq\frac16M\big(|x|+|y|\big)^3=\frac16e^{0.1}(0.2)^3=\frac{e^{0.1}}{750}\approx0.00147$$