経済数学演習問題 wLec 07, 2020/11/17
- I
-
$(x,y,z)$が
\begin{equation*}
\left\{
\begin{array}{ccc}
x-2y+z+1&=&0\\
x+y-z-2&=&0\\
\end{array}
\right.
\end{equation*}
を動くとき
\begin{equation*}
f(x,y,z)=x^2+y^2+z^2
\end{equation*}
の極値を求めましょう.可能ならば最小値を求めましょう.
- II
- 次の制約条件付の極値問題の停留点を求めましょう.
- (1)
\begin{equation*}
\left\{
\begin{array}{cccc}
g_1(x,y,z)&=&y^2+z^2-1&=0\\
g_2(x,y,z)&=&xz-3&=0
\end{array}
\right.
\end{equation*}
の下で
\begin{equation*}
w=f(x,y,z)=yz+zx
\end{equation*}
- (2)
\begin{equation*}
\left\{
\begin{array}{cccc}
g_1(x,y,z)&=&3x+y+z-5&=0\\
g_2(x,y,z)&=&x+y+z-1&=0
\end{array}
\right.
\end{equation*}
の下で
\begin{equation*}
w=f(x,y,z)=x^2+y^2+z^2
\end{equation*}
- (3)
\begin{equation*}
\left\{
\begin{array}{cccc}
g_1(x,y,z)&=&x^2+y^2+z^2-1&=0\\
g_2(x,y,z)&=y&=0
\end{array}
\right.
\end{equation*}
の下で
\begin{equation*}
w=f(x,y,z)=x+y+z^2
\end{equation*}
- III
- 以下の曲線の$\mathrm{P}_0$における接線の方向ベクトルを
求めましょう.
- (1)
\begin{equation*}
\left\{
\begin{array}{ccccc}
g_1(x,y,z)&=&x^2+y^2+z^2&=&0\\
g_2(x,y,z)&=&x-y+2z&=&0
\end{array}
\right.
\end{equation*}
at $\mathrm{P}_0(-\frac 1{\sqrt 3},\frac 1{\sqrt 3},\frac 1{\sqrt 3})$.
- (2)
\begin{equation*}
\left\{
\begin{array}{ccccc}
g_1(x,y,z)&=&x^2-y^2-z^2-1&=&0\\
g_2(x,y,z)&=&x+y+z&=&0
\end{array}
\right.
\end{equation*}
at $\mathrm{P}_0(\frac 2{\sqrt 2},-\frac 1{\sqrt 2},-\frac 1{\sqrt 2})$.