Functions of Several Variables: Very Difficult Problems with Solutions

Solution:
The function [tex]f[/tex] is differentiable everywhere, so its extrema can only be at stationary points:
[tex]f_x=4x^{3}-2x-2y=0,\quad f_y=4y^{3}-2x-2y=0[/tex]
Subtracting, [tex]4\left(x^{3}-y^{3}\right)=0[/tex], so [tex]x=y[/tex]; then [tex]4x^{3}-4x=0[/tex], and the stationary points are [tex](0, 0)[/tex], [tex](1, 1)[/tex], [tex](-1, -1)[/tex]. The second derivatives are
[tex]f_{xx}=12x^{2}-2,\quad f_{xy}=-2,\quad f_{yy}=12y^{2}-2[/tex]
At [tex](1, 1)[/tex] and [tex](-1, -1)[/tex]:
[tex]f_{xx}=f_{yy}=10,\quad \Delta=f_{xx}f_{yy}-f_{xy}^{2}=100-4=96\gt 0,\quad f_{xx}\gt 0[/tex]
so both points are local minima (with value [tex]-2[/tex]). At [tex](0, 0)[/tex] we have [tex]f_{xx}=f_{yy}=f_{xy}=-2[/tex] and [tex]\Delta=4-4=0[/tex], so the second-derivative test gives no answer, and we study [tex]f[/tex] near the origin directly; [tex]f(0,0)=0[/tex]. On the line [tex]y=x[/tex], for [tex]0\lt|x|\lt\sqrt{2}[/tex]:
[tex]f(x,x)=2x^{4}-4x^{2}=2x^{2}\left(x^{2}-2\right)\lt 0[/tex]
On the line [tex]y=-x[/tex], for [tex]x\ne 0[/tex]:
[tex]f(x,-x)=2x^{4}\gt 0[/tex]
Thus every neighborhood of the origin contains points where [tex]f\gt f(0,0)[/tex] and points where [tex]f\lt f(0,0)[/tex], so [tex](0, 0)[/tex] is not an extremum. The function has exactly [tex]2[/tex] points of local extremum.

Solution:
Put [tex]s=x-2y[/tex]. The partial derivatives are
[tex]u'_x=e^{-(x^2+y^2)}\left(1-2xs\right),\quad u'_y=e^{-(x^2+y^2)}\left(-2-2ys\right)[/tex]
At a stationary point [tex]2xs=1[/tex] and [tex]ys=-1[/tex]. Hence [tex]s\ne 0[/tex], [tex]x=\frac{1}{2s}[/tex], [tex]y=-\frac{1}{s}[/tex], and
[tex]s=x-2y=\frac{1}{2s}+\frac{2}{s}=\frac{5}{2s}\ \Rightarrow\ s^2=\frac{5}{2},\quad s=\pm\frac{\sqrt{10}}{2}[/tex]
This gives the stationary points [tex]P_1\left(\frac{\sqrt{10}}{10}, -\frac{\sqrt{10}}{5}\right)[/tex] (for [tex]s\gt 0[/tex]) and [tex]P_2\left(-\frac{\sqrt{10}}{10}, \frac{\sqrt{10}}{5}\right)[/tex]; at both of them [tex]x^2+y^2=\frac{1}{10}+\frac{4}{10}=\frac{1}{2}[/tex]. At a stationary point the terms containing [tex]1-2xs[/tex] and [tex]-2-2ys[/tex] vanish, so the second derivatives there are
[tex]u''_{xx}=-2(s+x)e^{-1/2},\quad u''_{xy}=4x\,e^{-1/2},\quad u''_{yy}=(4y-2s)e^{-1/2}[/tex]
At [tex]P_1[/tex], where [tex]s=\frac{\sqrt{10}}{2}[/tex]:
[tex]u''_{xx}=-\frac{6\sqrt{10}}{5}e^{-1/2}\lt 0,\quad u''_{xy}=\frac{2\sqrt{10}}{5}e^{-1/2},\quad u''_{yy}=-\frac{9\sqrt{10}}{5}e^{-1/2}[/tex]
[tex]u''_{xx}u''_{yy}-\left(u''_{xy}\right)^2=\frac{540-40}{25}e^{-1}=\frac{20}{e}\gt 0[/tex]
so [tex]P_1[/tex] is a local maximum:
[tex]u_{\max}=u(P_1)=\frac{\sqrt{10}}{2}e^{-1/2}=\sqrt{\frac{5}{2e}}[/tex]
Since [tex]u(-x,-y)=-u(x,y)[/tex] and [tex]P_2=-P_1[/tex], all second derivatives change sign at [tex]P_2[/tex]: the expression [tex]u''_{xx}u''_{yy}-\left(u''_{xy}\right)^2[/tex] is again [tex]\frac{20}{e}\gt 0[/tex], and now [tex]u''_{xx}\gt 0[/tex], so [tex]P_2[/tex] is a local minimum:
[tex]u_{\min}=u(P_2)=-\sqrt{\frac{5}{2e}}[/tex]

Solution:
For [tex](x,y)\ne(0,0)[/tex] we have [tex]f(x,y)=1+g(x,y)[/tex], where
[tex]g(x,y)=\frac{xy(x+y)}{x^2+y^2}[/tex]
Continuity. Since [tex]|xy|\le\frac{x^2+y^2}{2}[/tex], we get
[tex]|g(x,y)|\le\frac{|x+y|}{2}\le\frac{|x|+|y|}{2}[/tex]
and the right-hand side tends to [tex]0[/tex] as [tex](x,y)\to(0,0)[/tex]. Hence
[tex]\lim_{(x,y)\to(0,0)}f(x,y)=1=f(0,0)[/tex]
so [tex]f[/tex] is continuous at [tex]O[/tex]. Partial derivatives at [tex]O[/tex]: [tex]g(x,0)=g(0,y)=0[/tex], so [tex]f(x,0)=f(0,y)=1[/tex] and
[tex]f'_x(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=0,\quad f'_y(0,0)=\lim_{h\to 0}\frac{f(0,h)-f(0,0)}{h}=0[/tex]
Differentiability. If [tex]f[/tex] were differentiable at [tex]O[/tex], its differential there would be [tex]f'_x(0,0)\,dx+f'_y(0,0)\,dy=0[/tex], and we would have
[tex]\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}=\lim_{(x,y)\to(0,0)}\frac{g(x,y)}{\sqrt{x^2+y^2}}=0[/tex]
But along the ray [tex]y=x[/tex], [tex]x\gt 0[/tex]:
[tex]g(x,x)=\frac{x\times x\times 2x}{2x^2}=x,\qquad \frac{g(x,x)}{\sqrt{2x^2}}=\frac{x}{\sqrt{2}\,x}=\frac{1}{\sqrt{2}}\ne 0[/tex]
Hence the limit is not [tex]0[/tex], and [tex]f[/tex] is not differentiable at [tex]O[/tex], although it is continuous there and has both partial derivatives.

Solution:
The gradient [tex](2x, 2y, 0)[/tex] of the constraint function does not vanish on the cylinder [tex]x^{2}+y^{2}=1[/tex], so we may use the Lagrange function
[tex]L=x+y+z^{2}-\lambda\left(x^{2}+y^{2}-1\right)[/tex]
Its stationary points satisfy
[tex]L_x=1-2\lambda x=0,\quad L_y=1-2\lambda y=0,\quad L_z=2z=0,\quad x^{2}+y^{2}=1[/tex]
Hence [tex]z=0[/tex], [tex]x=y=\frac{1}{2\lambda}[/tex] and [tex]2x^{2}=1[/tex], which gives [tex]M_1\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0\right)[/tex] with [tex]\lambda=\frac{\sqrt{2}}{2}[/tex] and [tex]M_2\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}, 0\right)[/tex] with [tex]\lambda=-\frac{\sqrt{2}}{2}[/tex]. The second differential of [tex]L[/tex] is
[tex]d^{2}L=-2\lambda\left(dx^{2}+dy^{2}\right)+2\,dz^{2}[/tex]
and it must be considered on the tangent space [tex]x\,dx+y\,dy=0[/tex], i.e. for [tex]dy=-dx[/tex] at both points. At [tex]M_1[/tex]:
[tex]d^{2}L=-2\sqrt{2}\,dx^{2}+2\,dz^{2}[/tex]
This form takes both signs (it is negative for [tex]dz=0[/tex] and positive for [tex]dx=0[/tex]), so [tex]M_1[/tex] is not a conditional extremum. At [tex]M_2[/tex]:
[tex]d^{2}L=2\sqrt{2}\,dx^{2}+2\,dz^{2}\gt 0[/tex]
for [tex](dx,dz)\ne(0,0)[/tex], so [tex]M_2[/tex] is a conditional minimum:
[tex]f_{\min}=f\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}, 0\right)=-\sqrt{2}[/tex]
Since [tex]M_1[/tex] was the only other candidate, there is no conditional maximum. Check: with [tex]x=\cos\theta[/tex], [tex]y=\sin \theta[/tex] we get [tex]f=\sqrt{2}\sin \left(\theta+\frac{\pi}{4}\right)+z^{2}\ge-\sqrt{2}[/tex], while [tex]f\to+\infty[/tex] as [tex]z\to\infty[/tex].

Solution:
Let [tex]F[/tex] be the left-hand side of the equation. Then
[tex]F'_x=2x-z+2,\quad F'_y=2y-z+2,\quad F'_z=2z-x-y+2[/tex]
and where [tex]F'_z\ne 0[/tex] we have [tex]z'_x=-\frac{F'_x}{F'_z}[/tex], [tex]z'_y=-\frac{F'_y}{F'_z}[/tex]. At a stationary point [tex]F'_x=F'_y=0[/tex], so [tex]z=2x+2=2y+2[/tex], i.e. [tex]y=x[/tex], [tex]z=2x+2[/tex]. Substituting into the given equation:
[tex]2x^2+(2x+2)^2-2x(2x+2)+4x+2(2x+2)-2=0\ \Rightarrow\ 2x^2+12x+6=0\ \Rightarrow\ x=-3\pm\sqrt{6}[/tex]
So the points are [tex]x=y=-3\pm\sqrt{6}[/tex], [tex]z=-4\pm 2\sqrt{6}[/tex], and there [tex]F'_z=2z-2x+2=2x+6=\pm 2\sqrt{6}\ne 0[/tex], so the implicit function theorem applies. Differentiating [tex]F'_x+F'_z\,z'_x=0[/tex] once more and using [tex]z'_x=z'_y=0[/tex] at these points, we get
[tex]z''_{xx}=-\frac{F''_{xx}}{F'_z}=-\frac{2}{F'_z},\quad z''_{yy}=-\frac{2}{F'_z},\quad z''_{xy}=-\frac{F''_{xy}}{F'_z}=0[/tex]
[tex]z''_{xx}z''_{yy}-\left(z''_{xy}\right)^2=\frac{4}{(F'_z)^2}\gt 0[/tex]
so both points are extrema. For [tex]x=-3+\sqrt{6}[/tex]: [tex]F'_z=2\sqrt{6}\gt 0[/tex] and [tex]z''_{xx}=-\frac{1}{\sqrt{6}}\lt 0[/tex], a maximum:
[tex]z_{\max}=-4+2\sqrt{6}[/tex]
For [tex]x=-3-\sqrt{6}[/tex]: [tex]F'_z=-2\sqrt{6}\lt 0[/tex] and [tex]z''_{xx}=\frac{1}{\sqrt{6}}\gt 0[/tex], a minimum:
[tex]z_{\min}=-4-2\sqrt{6}[/tex]
Geometrically, the surface is an ellipsoid, and these are its highest and lowest points; they lie on the two different branches [tex]z(x,y)[/tex] obtained by solving the equation as a quadratic in [tex]z[/tex].

Solution:
We write [tex]z_x=\frac{\partial z}{\partial x}[/tex], [tex]z_{xx}=\frac{\partial^{2}z}{\partial x^{2}}[/tex], etc. Put [tex]t=ax+by+z[/tex] and [tex]F(x,y,z)=xy+z-f(t)[/tex]. Since [tex]F_z=1-f'(t)\ne 0[/tex], by the implicit function theorem [tex]z(x,y)[/tex] is twice differentiable, and we may differentiate the identity [tex]xy+z(x,y)=f(t)[/tex]. With respect to [tex]x[/tex] (note that [tex]t_x=a+z_x[/tex]):
[tex]y+z_x=f'(t)\left(a+z_x\right)\qquad(1)[/tex]
With respect to [tex]y[/tex] ([tex]t_y=b+z_y[/tex]):
[tex]x+z_y=f'(t)\left(b+z_y\right)\qquad(2)[/tex]
Differentiating [tex](1)[/tex] with respect to [tex]x[/tex]:
[tex]z_{xx}=f''(t)\left(a+z_x\right)^{2}+f'(t)\,z_{xx}\ \Rightarrow\ z_{xx}=\frac{f''(t)\left(a+z_x\right)^{2}}{1-f'(t)}\qquad(3)[/tex]
Differentiating [tex](2)[/tex] with respect to [tex]y[/tex] in the same way:
[tex]z_{yy}=\frac{f''(t)\left(b+z_y\right)^{2}}{1-f'(t)}\qquad(4)[/tex]
Now by [tex](3)[/tex] and [tex](2)[/tex]:
[tex]z_{xx}\left(z_y+x\right)^{2}=\frac{f''(t)\left(a+z_x\right)^{2}}{1-f'(t)}\times f'(t)^{2}\left(b+z_y\right)^{2}[/tex]
and by [tex](4)[/tex] and [tex](1)[/tex]:
[tex]z_{yy}\left(z_x+y\right)^{2}=\frac{f''(t)\left(b+z_y\right)^{2}}{1-f'(t)}\times f'(t)^{2}\left(a+z_x\right)^{2}[/tex]
The two right-hand sides are equal, which proves the identity.

Solution:
The distance from a point [tex](x,y,z)[/tex] to the plane is [tex]\frac{|Ax+By+Cz+1|}{\sqrt{A^{2}+B^{2}+C^{2}}}[/tex], and the required distance is the minimum of this expression over the ellipsoid (a compact set, so the minimum exists). First we find the range of [tex]\varphi=Ax+By+Cz[/tex] on the ellipsoid, using the Lagrange function
[tex]L=Ax+By+Cz-\mu\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}-1\right)[/tex]
[tex]L_x=A-\frac{2\mu x}{a^{2}}=0,\quad L_y=B-\frac{2\mu y}{b^{2}}=0,\quad L_z=C-\frac{2\mu z}{c^{2}}=0[/tex]
Since [tex](A,B,C)\ne(0,0,0)[/tex], we have [tex]\mu\ne 0[/tex], and
[tex]x=\frac{Aa^{2}}{2\mu},\quad y=\frac{Bb^{2}}{2\mu},\quad z=\frac{Cc^{2}}{2\mu}[/tex]
Substituting into the equation of the ellipsoid:
[tex]\frac{a^{2}A^{2}+b^{2}B^{2}+c^{2}C^{2}}{4\mu^{2}}=1\ \Rightarrow\ 2\mu=\pm K,\quad K=\sqrt{a^{2}A^{2}+b^{2}B^{2}+c^{2}C^{2}}[/tex]
and at these points
[tex]\varphi=\frac{a^{2}A^{2}+b^{2}B^{2}+c^{2}C^{2}}{2\mu}=\pm K[/tex]
The continuous function [tex]\varphi[/tex] attains its greatest and least values on the compact connected ellipsoid, necessarily at these points; hence [tex]\varphi[/tex] takes exactly the values in [tex][-K, K][/tex], and [tex]\varphi+1[/tex] those in [tex][1-K, 1+K][/tex]. The plane does not meet the ellipsoid if and only if [tex]0\notin[1-K, 1+K][/tex], i.e. [tex]K\lt 1[/tex]. Then [tex]\varphi+1\ge 1-K\gt 0[/tex] on the ellipsoid, and the least distance is
[tex]d_{\min}=\frac{1-K}{\sqrt{A^{2}+B^{2}+C^{2}}}=\frac{1-\sqrt{a^{2}A^{2}+b^{2}B^{2}+c^{2}C^{2}}}{\sqrt{A^{2}+B^{2}+C^{2}}}[/tex]