extreme (20)/r+5pir^2

Lösung

extrempunkte

\frac{20}{r} + 5 π r^{2}

Minimum (3 \frac{2}{π}, 10 \cdot 2^{\frac{2}{3}} 3 π + 5 π (\frac{2}{π})^{\frac{2}{3}})

Schritte zur Lösung

f^{'} (r) = - \frac{20}{r ^{2}} + 10 π r

Intervalle finden:

Absteigend

: - \infty < r < 0,

Absteigend

: 0 < r < \frac{3 4}{3 2 3 π},

Ansteigend

: \frac{3 4}{3 2 3 π} < r < \infty

Stecke

r = 3 \frac{2}{π}

\frac{20}{r} + 5 π r^{2} : 10 \cdot 2^{\frac{2}{3}} 3 π + 5 π (\frac{2}{π})^{\frac{2}{3}}

Minimum (3 \frac{2}{π}, 10 \cdot 2^{\frac{2}{3}} 3 π + 5 π (\frac{2}{π})^{\frac{2}{3}})

e x t re m e f (x) = 7 + x - x^{2} - ln ((x + 3)^{3})

e x t re m e f (x) = x^{6} + \frac{x}{6}

e x t re m e f (x) = x^{2} - 6 x y + 2 y^{2} + 10 x + 2 y - 5

e x t re m e 25 x

e x t re m e 2 - x^{4} + 2 x^{2} - y^{2}