A) 20
B) 16
C) 24
D) 8
E) 4
Correct Answer: D
Solution :
Given that, \[y=\frac{8-x}{2}\] Let \[p=xy=x\left( \frac{8-x}{2} \right)\] \[=\frac{8x-{{x}^{2}}}{2}\] On differentiating w.r.t.\[x,\]we get \[\frac{dp}{dx}=\frac{1}{2}(8-2x)\] Put \[\frac{dp}{dx}=0\] for maxima or minima \[\therefore \] \[8-2x=0\] \[\Rightarrow \] \[x=4\] Again \[\frac{{{d}^{2}}p}{d{{x}^{2}}}=\frac{1}{2}(-2)=-1\] \[{{\left( \frac{{{d}^{2}}p}{d{{x}^{2}}} \right)}_{x=4}}=-1<0\] Thus, function is maximum at\[x=4\]and\[y=2\] Therefore, maximum value of\[p=4\times 2=8\].
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