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 \[f(5)=25+50=75\].You need to login to perform this action.
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