1. Problem 17: Find the maximum and minimum values of the function $f(x,y)=x^3+3xy^2-15x^2-15y^2+72x$. 2. Compute partial derivatives and critical points. 3. $f_x=3x^2+3y^2-30x+72=3ig((x-5)^2+y^2-1ig)$, so $f_x=0\\ \Rightarrow (x-5)^2+y^2=1$. 4. $f_y=6xy-30y=6y(x-5)$, so $f_y=0\Rightarrow y=0\text{ or }x=5$. 5. Intersecting the conditions gives critical points $y=0$ and $(x-5)^2=1$ so $(6,0)$ and $(4,0)$, and $x=5$ with $y^2=1$ so $(5,1)$ and $(5,-1)$. 6. Evaluate $f$ at these points. 7. $f(6,0)=216-540+432=108$. 8. $f(4,0)=64-240+288=112$. 9. $f(5,1)=125+15-375-15+360=110$. 10. $f(5,-1)=110$. 11. Therefore the minimum value is $108$ at $(6,0)$ and the maximum value is $112$ at $(4,0)$. 1. Problem 18: Examine $f(x,y)=x^4+y^4-2(x-y)^2$ for extreme values. 2. Compute gradient: $f_x=4x^3-4(x-y)=4(x^3-x+y)$ and $f_y=4y^3+4(x-y)=4(y^3+y-x)$. 3. From $f_x-f_y=4(x-y)ig(x^2+xy+y^2-2\big)=0$ we get either $x=y$ or $x^2+xy+y^2=2$. 4. If $x=y$ then $4x^3=0$ so $x=y=0$ gives the critical point $(0,0)$. 5. Other critical points satisfy the nonlinear system $y=-x^3+x$ and $x^2+xy+y^2=2$, which may have additional real solutions found numerically, but analysis near the origin suffices to classify $(0,0)$. 6. Near $(0,0)$ the leading term is $-2(x-y)^2$ plus higher quartic terms, so directions with $x\neq y$ make $f$ decrease and along $x=y$ quartic positive behavior appears, hence $(0,0)$ is a saddle point rather than a local max or min. 1. Problem 19: Expand $e^x\cos y$ in a Taylor series about the point $(1,\pi/4)$. 2. Put $u=x-1$ and $v=y-\pi/4$ so $e^x\cos y=e^{1+u}\cos(\pi/4+v)=e\,e^u\big(\tfrac{1}{\sqrt{2}}(\cos v-\sin v)\big)$. 3. Expand $e^u=1+u+\tfrac{u^2}{2}+\tfrac{u^3}{6}+\cdots$, $\cos v=1-\tfrac{v^2}{2}+\tfrac{v^4}{24}+\cdots$, $\sin v=v-\tfrac{v^3}{6}+\cdots$. 4. Multiply to get the Taylor series; up to second order in $u,v$: 5. $$e^x\cos y\approx e\tfrac{1}{\sqrt{2}}\Big(1+u- v+\tfrac{u^2}{2}-\tfrac{v^2}{2}-uv\Big)\,.$$ 6. This gives the constant and low-order terms and the same method yields higher-order terms if needed. 1. Problem 20: Expand $g(x,y)=xy^2+\cos(xy)$ about $(1,\pi/2)$ up to terms containing second derivatives. 2. Let $u=x-1$ and $v=y-\pi/2$ and compute values and derivatives at $(1,\pi/2)$. 3. $g(1,\pi/2)=1\cdot(\tfrac{\pi}{2})^2+\cos(\tfrac{\pi}{2})=\tfrac{\pi^2}{4}+0$. 4. $g_x=y^2-y\sin(xy)$ so $g_x(1,\tfrac{\pi}{2})=\tfrac{\pi^2}{4}-\tfrac{\pi}{2}$. 5. $g_y=2xy-x\sin(xy)$ so $g_y(1,\tfrac{\pi}{2})=\pi-1$. 6. Second derivatives: $g_{xx}=-y^2\cos(xy)$ so $g_{xx}(1,\tfrac{\pi}{2})=0$, $g_{yy}=2x-x^2\cos(xy)$ so $g_{yy}(1,\tfrac{\pi}{2})=2$, $g_{xy}=2y-\sin(xy)-xy\cos(xy)$ so $g_{xy}(1,\tfrac{\pi}{2})=\pi-1$. 7. The second-order Taylor polynomial is 8. $$g\approx \tfrac{\pi^2}{4}+(\tfrac{\pi^2}{4}-\tfrac{\pi}{2})u+(\pi-1)v+\tfrac{1}{2}\big(0\cdot u^2+2(\pi-1)uv+2v^2\big)\,,$$ 9. which simplifies to the expansion shown above up to second-order terms. 1. Problem 21: Obtain the Taylor series expansion of $y^x$ about $(1,1)$ up to second degree terms and hence find the reciprocal near $(1,1)$. 2. Write $f(x,y)=e^{x\ln y}$ and let $u=x-1$, $v=y-1$. 3. At $(1,1)$ we have $f(1,1)=1$, $f_x(1,1)=0$, $f_y(1,1)=1$, $f_{xx}(1,1)=0$, $f_{xy}(1,1)=1$, $f_{yy}(1,1)=0$. 4. The second-degree Taylor polynomial is 5. $$y^x\approx 1+v+uv\,=1+(y-1)+(x-1)(y-1)\,.$$ 6. Hence the reciprocal up to second-degree terms is 7. $$\frac{1}{y^x}\approx 1-(v+uv)+v^2=1-(y-1)-(x-1)(y-1)+(y-1)^2\,,$$ 8. which is the expansion of $(y^x)^{-1}$ near $(1,1)$ up to second-degree terms. 1. Problem 22: Expand $h(x,y)=xy^2+x^2y$ in powers of $(x-1)$ and $(y+3)$ up to second degree terms. 2. Center at $(1,-3)$ and set $u=x-1$, $w=y+3$. 3. Evaluate and derivatives at $(1,-3)$: $h(1,-3)=6$, $h_x= y^2+2xy\Rightarrow h_x(1,-3)=3$, $h_y=2xy+x^2\Rightarrow h_y(1,-3)=-5$. 4. Second derivatives: $h_{xx}=2y\Rightarrow -6$, $h_{yy}=2x\Rightarrow 2$, $h_{xy}=2y+2x\Rightarrow -4$ at the point. 5. The second-order expansion is 6. $$h\approx 6+3u-5w-3u^2-4uw+w^2\,,$$ 7. where $u=x-1$ and $w=y+3$. 1. Problem 23: Expand $-\tan^{-1}(y/x)$ about $(1,1)$ up to third-degree terms. 2. Put $u=x-1$, $v=y-1$ and set $t=\dfrac{y}{x}$ and $w=t-1$ so $w=\dfrac{1+v}{1+u}-1$. 3. Expand $w$ up to third degree: $w\approx v-u+u^2-uv-u^3+u^2v$. 4. Use the one-variable Taylor expansion about $t=1$: 5. $$\arctan(1+w)\approx \tfrac{\pi}{4}+\tfrac{1}{2}w-\tfrac{1}{4}w^2+\tfrac{1}{12}w^3\,,$$ 6. hence 7. $$-\tan^{-1}\big(\tfrac{y}{x}\big)\approx -\tfrac{\pi}{4}-\tfrac{1}{2}w+\tfrac{1}{4}w^2-\tfrac{1}{12}w^3\,,$$ 8. substituting $w$ and collecting terms up to total degree three yields 9. $$-\tan^{-1}\big(\tfrac{y}{x}\big)\approx -\tfrac{\pi}{4}+\tfrac{1}{2}(x-1)-\tfrac{1}{2}(y-1)-\tfrac{1}{4}(x-1)^2+\tfrac{1}{4}(y-1)^2\\+\tfrac{1}{12}(x-1)^3-\tfrac{1}{4}(x-1)^2(y-1)-\tfrac{1}{4}(x-1)(y-1)^2-\tfrac{1}{12}(y-1)^3\,,$$ 10. which is the cubic expansion in $u,v$ about $(1,1)$. 1. Problem 25: Expand $e^x\log(1+y)$ in powers of $x,y$ up to third-degree terms. 2. Use $e^x=1+x+\tfrac{x^2}{2}+\tfrac{x^3}{6}+\cdots$ and $\log(1+y)=y-\tfrac{y^2}{2}+\tfrac{y^3}{3}+\cdots$ and multiply, keeping total degree $\le 3$. 3. The product up to third degree is 4. $$e^x\log(1+y)\approx y-\tfrac{y^2}{2}+\tfrac{y^3}{3}+xy-\tfrac{xy^2}{2}+\tfrac{x^2y}{2}\,.$$ 5. These are the terms through total degree three in $x,y$. Final answers summary: 17. min $108$ at $(6,0)$ and max $112$ at $(4,0)$. 18. $(0,0)$ is a saddle point and additional critical points satisfy a nonlinear system to be solved numerically. 19. $e^x\cos y\approx e\tfrac{1}{\sqrt{2}}\big(1+(x-1)-(y-\tfrac{\pi}{4})+\tfrac{(x-1)^2}{2}-\tfrac{(y-\tfrac{\pi}{4})^2}{2}-(x-1)(y-\tfrac{\pi}{4})\big)$ up to second order. 20. $xy^2+\cos(xy)\approx \tfrac{\pi^2}{4}+(\tfrac{\pi^2}{4}-\tfrac{\pi}{2})(x-1)+(\pi-1)(y-\tfrac{\pi}{2})+\tfrac{1}{2}\big(2(\pi-1)(x-1)(y-\tfrac{\pi}{2})+2(y-\tfrac{\pi}{2})^2\big)$. 21. $y^x\approx 1+(y-1)+(x-1)(y-1)$ and its reciprocal $\approx 1-(y-1)-(x-1)(y-1)+(y-1)^2$. 22. $xy^2+x^2y\approx 6+3(x-1)-5(y+3)-3(x-1)^2-4(x-1)(y+3)+(y+3)^2$. 23. $-\tan^{-1}(y/x)$ expansion about $(1,1)$ up to cubic terms is given above with $u=x-1$, $v=y-1$. 25. $e^x\log(1+y)\approx y-\tfrac{y^2}{2}+\tfrac{y^3}{3}+xy-\tfrac{xy^2}{2}+\tfrac{x^2y}{2}$ up to total degree three.