What is the extreme x^2+(405000)/x ?

The extreme x^2+(405000)/x is Minimum(15\sqrt[3]{60},225*60^{2/3}+(1800*15^{2/3})/(\sqrt[3]{4)})

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extreme x^2+(405000)/x

extreme

Minimum (15360, 225 \cdot 6 0^{\frac{2}{3}} + \frac{1800 \cdot 1 5 ^{\frac{2}{3}}}{3 4})

Solution steps

f^{'} (x) = 2 x - \frac{405000}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 15360,

Increasing

: 15360 < x < \infty

Plug

x = 15360

into

x^{2} + \frac{405000}{x} : 225 \cdot 6 0^{\frac{2}{3}} + \frac{1800 \cdot 1 5 ^{\frac{2}{3}}}{3 4}

Minimum (15360, 225 \cdot 6 0^{\frac{2}{3}} + \frac{1800 \cdot 1 5 ^{\frac{2}{3}}}{3 4})

What is the extreme x^2+(405000)/x ?
The extreme x^2+(405000)/x is Minimum(15\sqrt[3]{60},225*60^{2/3}+(1800*15^{2/3})/(\sqrt[3]{4)})