The minimum value of $\left(x^2+\frac{250}{x}\right)$ is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 75
$y=x^2+\frac{250}{x}⇒\frac{dy}{dx}=2x-\frac{250}{x^2}=0$
$⇒x=5$
$\frac{d^2y}{dx^2}=2>0⇒x=5$ (point of local min.)
so $y_{min} = y(5)=25+50=75$