A) \[-2\]
B) \[0\]
C) \[0\]
D) \[-3\]
Correct Answer: A
Solution :
Given that, \[x-2y=4\] Let \[A=xy\] \[\therefore \] \[A=x\left( \frac{x-4}{2} \right)\] \[=\frac{1}{2}({{x}^{2}}-4x)\] On differentiating w.r.t.\[x,\] we get \[\frac{dA}{dx}=\frac{1}{2}(2x-4)\] For maxima or minima, put\[2x-4=0\] \[\therefore \] \[x=2\] Again differentiating, we get \[\frac{{{d}^{2}}A}{d{{x}^{2}}}=\frac{1}{2}(2)=1\] At \[x=2,\,\,\frac{{{d}^{2}}A}{d{{x}^{2}}}=1\], minima \[\therefore \]The minimum value of \[A=xy=\frac{1}{2}(4-8)\] Note: Minimum and maximum values of a curve comes alternatively.
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