\input a4
\nopagenumbers
\input pstricks
\input fp
\input vec2d
\def\frac#1#2{{#1 \over #2}}

\VDDdef O(0,0)U(-1,-.5)V(7,5)X(1,0)Y(0,1)
\VDDdeflinepoint x:OX
\VDDdeflinepoint y:OY
\VDDdcanvas1:1
  \VDDdwholeline xy
  \VDDllineedgeletter y:V:-1:-1
  \VDDllineedgeletter x:V:1:1
  \VDDllinelinepointletter xy:OV:1:-1
  \VDDdef A(3,2.25)
  \VDDdef A'(2.25,2.85)
  \VDDdeflinepoint k:AA'
  \VDDdwholeline k
  \VDDdline AA''(3,0)
  \VDDdline AA'''(0,2.25)
  \VDDllinepointany x:A'':V:1:$u$
  \VDDllinepointany y:A''':V:1:$v$
  \VDDllinepointany k:A':V:1:$k$
  \VDDllinepointany k:A:V:-1:$A$
  \VDDcintersect P:kx
  \VDDcintersect Q:ky
  \VDDllinelinepointletter xk:PV:1:1
  \VDDllinelinepointany xk:PV:-1:1:$\alpha$
  \VDDllinelinepointletter yk:QV:1:1
\VDDdcanvasend

We berekenen de minimale lengte van $PQ$ exact. Er geldt $A(u, v)$.

$$PQ(\alpha) = \frac{v}{\sin(\alpha)} + \frac{u}{\cos(\alpha)}$$

$$PQ'(\alpha) = \frac{v\cos(\alpha)}{\sin^2(\alpha)} - \frac{u\sin(\alpha)}{\cos^2(\alpha)}$$

$L' = 0$ geeft $\tan(\alpha) = \root3\of{\frac{v}{u}} = \frac1u\root3\of{u^2v}$. Omdat $0 < \alpha < \frac12\pi$ geldt er dat $\cos(\alpha) = \sqrt{1 - \sin^2(\alpha)}$. Dit geeft
$$\frac{\sin(\alpha)}{\sqrt{1 - \sin^2(\alpha)}} = \frac{\root3\of{u^2v}}u,$$
ofwel
$$
\frac1{\sin^2(\alpha)} = \frac{\root3\of{uv^2} + u}{\root3\of{uv^2}}\cdot\frac{\root3\of{u^2v}}{\root3\of{u^2v}} = \frac{\root3\of{u^2v}+v}v.
%= \frac{16\root3\of{36} -24\root3\of{6}+ 36}{64+36} = \frac1{25}(9 - 6\root3\of6 + 4\root3\of{36}).
$$
Evenzo
$$\frac{\sqrt{1 - \cos^2(\alpha)}}{\cos(\alpha)} = \frac{\root3\of{u^2v}}u$$
geeft
$$\frac1{\cos^2(\alpha)} = \frac{\root3\of{uv^2}+u}u.$$

Dus
$$
%L_{\rm min} = 3\sqrt{\frac{2\root3\of6+3}3} + 4\sqrt{\frac{\root3\of{36}+4}4} = \sqrt{6\root3\of6+9} +2\sqrt{\root3\of{36}+4}
L_{\rm min} = v\sqrt{\frac{\root3\of{u^2v}+v}v} + u\sqrt{\frac{\root3\of{uv^2}+u}u} = \sqrt{v\root3\of{u^2v}+v^2} + \sqrt{u\root3\of{uv^2}+u^2}
%\sqrt{6\root3\of6+9} +2\sqrt{\root3\of{36}+4}
$$
en met $u = 4 \wedge v = 3$ hebben we
$$
L_{\rm min} = \sqrt{6\root3\of6+9} +2\sqrt{\root3\of{36}+4}.
$$

%$$
%L_{\rm min} = \frac{15\sqrt{9 - 6\root3\of6 + 4\root3\of{36}}}{9 - 6\root3\of6 + 4\root3\of{36}} + \frac{10\sqrt{16 + 6\root3\of6 - 4\root3\of{36}}}{8 + 3\root3\of6 - 2\root3\of{36}}
%$$
%$$
%L_{\rm min} = \frac{15\sqrt{9 - 6\root3\of6 + 4\root3\of{36}}}{9 - 6\root3\of6 + 4\root3\of{36}}\cdot\frac{8 + 3\root3\of6 - 2\root3\of{36}}{8 + 3\root3\of6 - 2\root3\of{36}} + \frac{10\sqrt{16 + 6\root3\of6 - 4\root3\of{36}}}{8 + 3\root3\of6 - 2\root3\of{36}}\cdot\frac{9 - 6\root3\of6 + 4\root3\of{36}}{9 - 6\root3\of6 + 4\root3\of{36}}
%$$
%$$
%L_{\rm min} = \frac{15\sqrt{9 - 6\root3\of6 + 4\root3\of{36}}\cdot(8 + 3\root3\of6 - 2\root3\of{36}) + 10\sqrt{16 + 6\root3\of6 - 4\root3\of{36}}\cdot(9 - 6\root3\of6 + 4\root3\of{36})}{216-69\root3\of6-4\root3\of{36}}
%$$
%boven en onder vermenigvuldigen met $24+8\root3\of6+3\root3\of{36}$ geeft
%$$
%L_{\rm min} =\frac15\left( (3+2\root3\of6)\sqrt{9 - 6\root3\of6 + 4\root3\of{36}} + (4+\root3\of{36})\sqrt{16 + 6\root3\of6 - 4\root3\of{36}}\right)
%$$
\bye
