The minimum value of $\left(x^2+\frac{25

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

The minimum value of $\left(x^2+\frac{250}{x}\right)$ is :

Options:

Correct Answer:

Explanation:

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 $y_{min} = y(5)=25+50=75$