극값과 안장점

a) (i) ${ \boldsymbol{\nabla} } f$ is zero at critical points. The point $T$ is a local maximum and the point $U$ is a saddle point. The remaining points $P$, $R$, $S$, are not critical points.

(a) (ii) Only $U$ is a saddle point.

(a) (iii) We have $f_y(x,y)>0$ if $f$ increases as you move vertically upward through $(x,y)$. Looking at the diagram, we see

$$\begin{aligned} f_y(P) <0\qquad f_y(Q) <0\qquad f_y(R) =0\qquad f_y(S) >0\qquad f_y(T) =0\qquad f_y(U) =0 \end{aligned}$$

So only $S$ works.

(a) (iv) The directional derivative of $f$ in the direction $\left< 0,-1\right>$ is ${ \boldsymbol{\nabla} } f\cdot \left< 0,-1\right> =-f_y$. It is negative if and only if $f_y>0$. So, again, only $S$ works.

(b) (i) The function $z=F(x,2)$ is increasing at $x=1$, because the $y=2.0$ graph in the diagram has positive slope at $x=1$. So $F_x(1,2)>0$.

(b) (ii) The function $z=F(x,2)$ is also increasing (though slowly) at $x=2$, because the $y=2.0$ graph in the diagram has positive slope at $x=2$. So $F_x(2,2)>0$. So $F$ does not have a critical point at $(2,2)$.

(b) (iii) From the diagram the looks like $F_x(1,1.9) > F_x(1,2.0) > F_x(1,2.1)$. That is, it looks like the slope of the $y=1.9$ graph at $x=1$ is larger than the slope of the $y=2.0$ graph at $x=1$, which in turn is larger than the slope of the $y=2.1$ graph at $x=1$. So it looks like $F_x(1,y)$ decreases as $y$ increases through $y=2$, and consequently $F_{xy}(1,2)<0$.

The height $\sqrt{x^2+y^2}$ at $(x,y)$ is the distance from $(x,y)$ to $(0,0)$. So the minimum height is zero at $(0,0,0)$. The surface is a cone. The cone has a point at $(0,0,0)$ and the derivatives $z_x$ and $z_y$ do not exist there. The maximum height is achieved when $(x,y)$ is as far as possible from $(0,0)$. The highest points are at $(\pm 1,\pm 1,\sqrt{2})$. There $z_x$ and $z_y$ exist but are not zero. These points would not be the highest points if it were not for the restriction $|x|,|y|\le 1$.

Define $f(t)=g(\mathbf{a}+t\mathbf{d})$ and determine $t_0$ by $\mathbf{x}_0=\mathbf{a}+t_0\mathbf{d}$. Then $f'(t)={ \boldsymbol{\nabla} } g(\mathbf{a}+t\mathbf{d})\cdot\mathbf{d}$. To see this, write $\mathbf{a}=\left< a_1,a_2,a_3\right>$ and $\mathbf{d}=\left< d_1,d_2,d_3\right>$. Then

$$f(t)=g(a_1+td_1,a_2+td_2,a_3+td_3)$$

So, by the chain rule,

$$\begin{aligned} f'(t) &= \frac{\partial g}{\partial x}(a_1+td_1,a_2+td_2,a_3+td_3)\,d_1 +\frac{\partial g}{\partial y}(a_1+td_1,a_2+td_2,a_3+td_3)\,d_2 \\& +\frac{\partial g}{\partial z}(a_1+td_1,a_2+td_2,a_3+td_3)\,d_3 \\ &={ \boldsymbol{\nabla} } g(\mathbf{a}+t\mathbf{d})\cdot\mathbf{d} \end{aligned}$$

Then $\mathbf{x}_0$ is a local max or min of the restriction of $g$ to the specified line if and only if $t_0$ is a local max or min of $f(t)$. If so, $f'(t_0)$ necessarily vanishes. So if $\mathbf{x}_0$ is a local max or min of the restriction of $g$ to the specified line, then ${ \boldsymbol{\nabla} } g(\mathbf{x}_0)\cdot\mathbf{d}=0$, i.e. ${ \boldsymbol{\nabla} } g(\mathbf{x}_0)\perp\mathbf{d}$, and $\mathbf{x}_0=\mathbf{a}+t_0\mathbf{d}$ for some $t_0$. The second condition is to ensure that $x_0$ lies on the line.