Zentrale Abiturprüfung Berlin 2008 Lösunsansatz

Hi,

erste Ansätze sind hier folgende:

allg. Volumenformel Zylinder:

${\displaystyle \mathrm{V<mspace\; width="0.12em"></mspace>}=\mathrm{\pi <mspace\; width="0.12em"></mspace>}\cdot {\mathrm{r<mspace\; width="0.12em"></mspace>}}^2\cdot \mathrm{h<mspace\; width="0.12em"></mspace>}}$

aus der Skizze folgt:

${\displaystyle \mathrm{r<mspace\; width="0.12em"></mspace>}=\mathrm{f<mspace\; width="0.12em"></mspace>}\left(\mathrm{a<mspace\; width="0.12em"></mspace>}\right)=\sqrt{2\mathrm{a<mspace\; width="0.12em"></mspace>}}=\mathrm{f<mspace\; width="0.12em"></mspace>}\left(\mathrm{b<mspace\; width="0.12em"></mspace>}\right)=\sqrt{4\mathrm{b<mspace\; width="0.12em"></mspace>}-{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2}\left(1\right)}$

${\displaystyle \mathrm{h<mspace\; width="0.12em"></mspace>}=\mathrm{b<mspace\; width="0.12em"></mspace>}-\mathrm{a<mspace\; width="0.12em"></mspace>}\left(2\right)}$

aus (1) folgt:

${\displaystyle \sqrt{2\mathrm{a<mspace\; width="0.12em"></mspace>}}=\sqrt{4\mathrm{b<mspace\; width="0.12em"></mspace>}-{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2}}$

${\displaystyle 2\mathrm{a<mspace\; width="0.12em"></mspace>}=4\mathrm{b<mspace\; width="0.12em"></mspace>}-{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2}$

${\displaystyle \mathrm{a<mspace\; width="0.12em"></mspace>}=\frac{4\mathrm{b<mspace\; width="0.12em"></mspace>}-{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2}{2}}$

mit (2) folgt dann:

${\displaystyle \mathrm{h<mspace\; width="0.12em"></mspace>}=\mathrm{b<mspace\; width="0.12em"></mspace>}-\left(\frac{4\mathrm{b<mspace\; width="0.12em"></mspace>}-{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2}{2}\right)=\frac{2\mathrm{b<mspace\; width="0.12em"></mspace>}-4\mathrm{b<mspace\; width="0.12em"></mspace>}+{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2}{2}=\frac{-2\mathrm{b<mspace\; width="0.12em"></mspace>}+{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2}{2}}$

in Volumenformel eingesetzt:

${\displaystyle \mathrm{V<mspace\; width="0.12em"></mspace>}\left(\mathrm{b<mspace\; width="0.12em"></mspace>}\right)=\mathrm{\pi <mspace\; width="0.12em"></mspace>}\cdot {\left(\sqrt{4\mathrm{b<mspace\; width="0.12em"></mspace>}-{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2}\right)}^2\cdot \left(\frac{-2\mathrm{b<mspace\; width="0.12em"></mspace>}+{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2}{2}\right)}$

${\displaystyle \mathrm{V<mspace\; width="0.12em"></mspace>}\left(\mathrm{b<mspace\; width="0.12em"></mspace>}\right)=\mathrm{\pi <mspace\; width="0.12em"></mspace>}\cdot (4\mathrm{b<mspace\; width="0.12em"></mspace>}-{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2)\left(\frac{-2\mathrm{b<mspace\; width="0.12em"></mspace>}+{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2}{2}\right)=\frac{\mathrm{\pi <mspace\; width="0.12em"></mspace>}}{2}(4\mathrm{b<mspace\; width="0.12em"></mspace>}-{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2)(-2\mathrm{b<mspace\; width="0.12em"></mspace>}+{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2)}$

${\displaystyle \mathrm{V<mspace\; width="0.12em"></mspace>}\left(\mathrm{b<mspace\; width="0.12em"></mspace>}\right)=\frac{\mathrm{\pi <mspace\; width="0.12em"></mspace>}}{2}(-8{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2+2{\mathrm{b<mspace\; width="0.12em"></mspace>}}^3+4{\mathrm{b<mspace\; width="0.12em"></mspace>}}^3-{\mathrm{b<mspace\; width="0.12em"></mspace>}}^4)}$

${\displaystyle \mathrm{V<mspace\; width="0.12em"></mspace>}\left(\mathrm{b<mspace\; width="0.12em"></mspace>}\right)=\frac{\mathrm{\pi <mspace\; width="0.12em"></mspace>}}{2}(-8{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2+6{\mathrm{b<mspace\; width="0.12em"></mspace>}}^3-{\mathrm{b<mspace\; width="0.12em"></mspace>}}^4)}$

${\displaystyle \mathrm{V<mspace\; width="0.12em"></mspace>}\left(\mathrm{b<mspace\; width="0.12em"></mspace>}\right)=\mathrm{\pi <mspace\; width="0.12em"></mspace>}(-4{\mathrm{b<mspace\; width="0.12em"></mspace>}}^2+3{\mathrm{b<mspace\; width="0.12em"></mspace>}}^3-\frac{1}{2}{\mathrm{b<mspace\; width="0.12em"></mspace>}}^4)}$

das wäre die Zielfunktion, d) schaffst du dann bestimmt selbst

Gruß pantau