A) \[1600\,{{m}^{2}}\]
B) \[2100\,{{m}^{2}}\]
C) \[2400\,{{m}^{2}}\]
D) \[2500\,{{m}^{2}}\]
Correct Answer: D
Solution :
[d] Let length and breadth of rectangular field be x and y respectively \[\therefore 2(x+y)=200\Rightarrow y=100-x\] and area, \[A=xy\] \[=x(100-x)\because \frac{dA}{dx}=100-2x\] Put \[\frac{dA}{dx}=0\] for maxima or minima \[100-2x=0\] \[\Rightarrow x=50\Rightarrow y=50\] Now, \[\frac{{{d}^{2}}A}{d{{x}^{2}}}=-2<0\], which shows maximum, independent of values of x and y, but only when they are equal. \[\therefore \] A is maximum at \[x=50\]. Hence, required area \[=50(100-50)\] \[=50\times 50=2500\,{{m}^{2}}\]
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