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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Maxima and Minima

  • question_answer
    The minimum value of \[\left( {{x}^{2}}+\frac{250}{x} \right)\] is [Kurukshetra CEE 2002]

    A)            75

    B)            50

    C)            25

    D)            55

    Correct Answer: A

    Solution :

               Let \[y=f(x)=\left( {{x}^{2}}+\frac{250}{x} \right)\],  \\[\frac{dy}{dx}={f}'(x)=2x-\frac{250}{{{x}^{2}}}\]            Put \[{f}'(x)=0\]Þ\[2{{x}^{3}}-250=0\]Þ\[{{x}^{3}}=125\]Þ \[x=5\]            Again,\[\frac{{{d}^{2}}y}{d{{x}^{2}}}={f}''(x)=2+\frac{500}{{{x}^{3}}}\].Now \[{f}''(5)=2+\frac{500}{125}>0\]            Hence at \[x=5\]. The function will be minimum.                    Minimum value \[f(5)=25+50=75\].


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