Im supposed to present it such that its min d, being the distance, subject to this and that, but while they said not to solve I only know how to solve it...
optimization of min/max from point C and circle C to a line
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optimization
1 Answers
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For (a) you can use half the distance squared to make calculations easier. Then using $y=(c-ax)/b$, we get $$\frac{1}{2}d^2(x,y)=\frac{1}{2}(x-x_c)^2+\frac{1}{2}(y-y_c)^2=\frac{1}{2}((x-x_c)^2+((c-a\,x)/b-y_c)^2)=\frac{1}{2}d^2(x).$$ We find the max/min of a function when it's derivative is zero w.r.t. the free variable, in this case $x$. So we solve $x$ for $$\frac{df}{dx}=(x-x_c)+((c-a\,x)/b-y_c)=x(1-\frac{a}{b})-x_c-y_c+\frac{c}{b}=0.$$ This, it turns out, is solved by $$x=\frac{a c + b^2 x_c - a b\,y_c}{a^2 + b^2}.$$ Recalling that $y=(c-a\,x)/b$, $$y=\frac{b c - a b\,x_c + a^2\,y_c}{a^2 + b^2}.$$
For b), we parametrize the circle as $\hat{r}(t)=\langle r\cos{t},r\sin{t}\rangle$. The value, $f(x,y)$, at $\hat{r}(t)$ is equal to $$f(r\cos{t},r\sin{t})=a\,r\cos{t}+b\,r\sin{t}=f(t).$$ As above, take the derivative w.r.t $t$ and set it to $0$: $$\frac{df}{dt}=-a\,r\sin{t}+b\,r\cos{t}=0\quad\Rightarrow\quad\frac{\sin{t}}{\cos{t}}=\frac{r\cdot b}{r\cdot a}=\frac{b}{a}\quad\Rightarrow\quad t=\arctan{\frac{b}{a}}.$$ Plugging back into $\hat{r}(t)$, we get $$\hat{r}\left(\arctan{\frac{b}{a}}\right)=\langle r\cos{\arctan{\frac{b}{a}}},r\sin{\arctan{\frac{b}{a}}}\rangle=\langle \frac{a r}{\sqrt{a^2+b^2}},\frac{rb}{\sqrt{a^2+b^2}}\rangle.$$