What is the extreme f(x)=(20)/x+2pix^2,x>0 ?

The extreme f(x)=(20)/x+2pix^2,x>0 is Minimum(\sqrt[3]{5/pi},4*5^{2/3}\sqrt[3]{pi}+2pi(5/pi)^{2/3})

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extreme f(x)=(20)/x+2pix^2,x>0

extreme points

f (x) = \frac{20}{x} + 2 π x^{2}, x > 0

Solution steps

Find the critical points:

Domain of

\frac{20}{x} + 2 π x^{2}, x > 0 : x > 0

Find intervals:

Decreasing

Increasing

Plug the extreme point

into

f (x) = - x^{3} + x^{2} + 11 x - 1

f (x) = 2 x - sin (x)

f (x) = \frac{x ^{2}}{1 + x}, - \frac{1}{2} \leq x \leq 1

f (x, y) = x^{3} + y^{3} - 300 x - 243 y + 2

f (x, y) = - 6 x^{2} - 7 x y - 6 y^{2} - 18 x - 58 y + 8

What is the extreme f(x)=(20)/x+2pix^2,x>0 ?
The extreme f(x)=(20)/x+2pix^2,x>0 is Minimum(\sqrt[3]{5/pi},4*5^{2/3}\sqrt[3]{pi}+2pi(5/pi)^{2/3})