Functions of Several Variables: Problems with Solutions

Solution:
The function [tex]\arcsin u[/tex] is defined only for [tex]-1\le u\le 1[/tex], so we need
[tex]-1\le 25x^2+9y^2\le 1[/tex]
The left inequality always holds, because [tex]25x^2+9y^2\ge 0\gt -1[/tex]. The right one can be written as
[tex]25x^2+9y^2\le 1\iff \frac{x^2}{\left(\frac{1}{5}\right)^2}+\frac{y^2}{\left(\frac{1}{3}\right)^2}\le 1[/tex]
So the domain is the ellipse with center [tex]O(0, 0)[/tex] and semi-axes [tex]\frac{1}{5}[/tex] (along the [tex]Ox[/tex] axis) and [tex]\frac{1}{3}[/tex] (along the [tex]Oy[/tex] axis) together with its interior. The ellipse itself belongs to the domain, because [tex]\arcsin 1[/tex] is defined:
[tex]D=\left\{(x,y):\ 25x^2+9y^2\le 1\right\}[/tex]

Solution:
Here [tex]z[/tex] depends on [tex]t[/tex] through [tex]x[/tex] and [tex]y[/tex]. The function [tex]e^{x-2y}[/tex] has continuous partial derivatives and [tex]x[/tex], [tex]y[/tex] are differentiable functions of [tex]t[/tex], so by the chain rule
[tex]\frac{dz}{dt}=z'_x\,\frac{dx}{dt}+z'_y\,\frac{dy}{dt}[/tex]
We have
[tex]z'_x=e^{x-2y},\quad z'_y=-2e^{x-2y},\quad \frac{dx}{dt}=\cos t,\quad \frac{dy}{dt}=3t^2[/tex]
hence
[tex]\frac{dz}{dt}=e^{x-2y}\cos t-2e^{x-2y}\times 3t^2=e^{\sin t-2t^3}\left(\cos t-6t^2\right)[/tex]
Check: directly [tex]z=e^{\sin t-2t^3}[/tex], and the derivative of the exponent is [tex]\cos t-6t^2[/tex], which gives the same result.

Solution:
For [tex]x\gt 0[/tex] the function [tex]z=x^y=e^{y\ln x}[/tex] has continuous partial derivatives of all orders. We differentiate with respect to [tex]x[/tex] ([tex]y[/tex] is a constant, power rule):
[tex]z'_x=y\,x^{y-1}[/tex]
Now we differentiate [tex]z'_x[/tex] with respect to [tex]y[/tex] ([tex]x[/tex] is a constant). Both factors depend on [tex]y[/tex], so we use the product rule and [tex]\left(x^{y-1}\right)'_y=x^{y-1}\ln x[/tex]:
[tex]z''_{xy}=1\times x^{y-1}+y\,x^{y-1}\ln x=x^{y-1}\left(1+y\ln x\right)[/tex]
Check in the other order (by Schwarz's theorem the result must be the same):
[tex]z'_y=x^y\ln x,\qquad z''_{yx}=y\,x^{y-1}\ln x+x^y\times \frac{1}{x}=x^{y-1}\left(1+y\ln x\right)[/tex]

Solution:
Put [tex]F(x,y,z)=x^3+2y^3+z^2-3xyz-2y+3[/tex]. The function [tex]F[/tex] has continuous partial derivatives, so at the points where [tex]F'_z\ne 0[/tex] the implicit function theorem gives
[tex]z'_y=-\frac{F'_y}{F'_z}[/tex]
Here
[tex]F'_y=6y^2-3xz-2,\qquad F'_z=2z-3xy[/tex]
Hence
[tex]z'_y=-\frac{6y^2-3xz-2}{2z-3xy}=\frac{3xz-6y^2+2}{2z-3xy}[/tex]
The same result is obtained by differentiating the identity with respect to [tex]y[/tex], keeping in mind that [tex]z=z(x,y)[/tex]:
[tex]6y^2+2z\,z'_y-3xz-3xy\,z'_y-2=0\ \Rightarrow\ z'_y\left(2z-3xy\right)=3xz-6y^2+2[/tex]

Solution:
The function has continuous partial derivatives near [tex]M[/tex] (there [tex]x^{2}+y^{2}\ne 0[/tex]), hence it is differentiable at [tex]M[/tex], and its derivative in the direction of a unit vector [tex]\vec{l}_0=\left(\cos\alpha, \cos\beta\right)[/tex] is
[tex]\frac{\partial z}{\partial l}(M)=z'_x(M)\cos\alpha+z'_y(M)\cos\beta[/tex]
Using [tex]\left(\operatorname{arccot}u\right)'=-\frac{1}{1+u^{2}}[/tex], we get
[tex]z'_x=-\frac{3}{1+(3x+4y)^{2}}+\frac{2x}{x^{2}+y^{2}},\qquad z'_y=-\frac{4}{1+(3x+4y)^{2}}+\frac{2y}{x^{2}+y^{2}}[/tex]
At [tex]M(-1, 1)[/tex] we have [tex]3x+4y=1[/tex] and [tex]x^{2}+y^{2}=2[/tex], so
[tex]z'_x(M)=-\frac{3}{2}-1=-\frac{5}{2},\qquad z'_y(M)=-2+1=-1[/tex]
Next, [tex]\overrightarrow{MN}=(4, 3)[/tex] and [tex]\left|\overrightarrow{MN}\right|=\sqrt{16+9}=5[/tex], so [tex]\vec{l}_0=\left(\frac{4}{5}, \frac{3}{5}\right)[/tex]. Therefore
[tex]\frac{\partial z}{\partial l}(M)=-\frac{5}{2}\times \frac{4}{5}+(-1)\times \frac{3}{5}=-2-\frac{3}{5}=-\frac{13}{5}[/tex]

Solution:
Take [tex]x_0=1[/tex], [tex]y_0=2[/tex], [tex]\Delta x=0.05[/tex], [tex]\Delta y=0.01[/tex]. The function has continuous partial derivatives, so it is differentiable and for small increments
[tex]z\left(x_0+\Delta x,\,y_0+\Delta y\right)\approx z\left(x_0,y_0\right)+dz,\qquad dz=z'_x\left(x_0,y_0\right)\Delta x+z'_y\left(x_0,y_0\right)\Delta y[/tex]
We compute
[tex]z(1, 2)=3-4-\frac{\sin 2\pi}{\pi}=-1[/tex]
[tex]z'_x=6x-y^2\ \Rightarrow\ z'_x(1, 2)=6-4=2[/tex]
[tex]z'_y=-2xy-\cos \pi y\ \Rightarrow\ z'_y(1, 2)=-4-\cos 2\pi=-5[/tex]
[tex]dz=2\times 0.05+(-5)\times 0.01=0.1-0.05=0.05[/tex]
Therefore
[tex]z\left(1.05, 2.01\right)\approx -1+0.05=-0.95[/tex]

Solution:
The profit is
[tex]P(x,y)=R(x,y)-C(x,y)=-2x^{2}+2xy-3y^{2}+900x+1800y[/tex]
Stationary points:
[tex]\begin{cases}P'_x=-4x+2y+900=0\\ P'_y=2x-6y+1800=0\end{cases}[/tex]
From the first equation [tex]y=2x-450[/tex]. Substituting into the second gives [tex]2x-12x+2700+1800=0[/tex], so [tex]x=450[/tex] and [tex]y=450[/tex]. The second derivatives are constant:
[tex]P''_{xx}=-4,\quad P''_{xy}=2,\quad P''_{yy}=-6,\qquad \Delta=P''_{xx}P''_{yy}-\left(P''_{xy}\right)^{2}=24-4=20\gt 0[/tex]
Since [tex]\Delta\gt 0[/tex] and [tex]P''_{xx}\lt 0[/tex], the point [tex](450, 450)[/tex] is a local maximum. Moreover, the second derivatives are the same everywhere, so [tex]P[/tex] is a concave function on the whole plane and this maximum is global. The quantities [tex]x=y=450[/tex] are whole numbers, hence admissible. The maximum profit is
[tex]P(450, 450)=-2\times 450^{2}+2\times 450^{2}-3\times 450^{2}+900\times 450+1800\times 450=-607500+405000+810000=607500[/tex]
The maximum profit is [tex]607500[/tex] dollars.