A=\( \frac{a}{2} \) *[ f(a)] soll maximal werden.
f(x)= \( \frac{1}{8} \) x^3-\( \frac{17}{8} \) x^2+\( \frac{35}{4} \) x
f(a)= \( \frac{1}{8} \) a^3 -\( \frac{17}{8} \) a^2+\( \frac{35}{4} \) a
A=\( \frac{a}{2} \) *( \( \frac{1}{8} \) a^3 -\( \frac{17}{8} \) a^2+\( \frac{35}{4} \) a)
A= \( \frac{1}{16} \) a^4 -\( \frac{17}{16} \) a^3+\( \frac{35}{8} \) a^2
A´= \( \frac{1}{4} \)a^3 -\( \frac{51}{16} \)a^2+\( \frac{35}{4} \) a
\( \frac{1}{4} \)a^3 -\( \frac{51}{16} \)a^2+\( \frac{35}{4} \) a=0
a*(\( \frac{1}{4} \)a^2 -\( \frac{51}{16} \)a+\( \frac{35}{4} \) )=0
a_1=0
\( \frac{1}{4} \)a^2 -\( \frac{51}{16} \)a+\( \frac{35}{4} \) =0|*4
a^2 -\( \frac{51}{4} \)a+35=0
a_2=4
a_3 =\( \frac{35}{4} \) → kommt nicht in Frage, weil außerhalb des Intervalls
Art des Extremwertes:
A´´= \( \frac{3}{4} \)a^2 -\( \frac{51}{8} \)a+\( \frac{35}{4} \)
A´´(0)= \( \frac{35}{4} \)>0→Minimum
A´´(4)= \( \frac{3}{4} \)*4^2 -\( \frac{51}{8} \)*4+\( \frac{35}{4} \)=12-25,5+8,75 <0→Maximum
A_max=...