A) \[2000\]
B) \[2500\]
C) \[3000\]
D) \[3500\]
Correct Answer: B
Solution :
Given, \[\frac{dP}{dx}=100-12\sqrt{x}\] \[\Rightarrow \] \[dP=(100-12\sqrt{x})dx\] On integrating both sides, we get \[P=100x-12\left( \frac{{{x}^{3/2}}}{3/2} \right)+C\] \[\Rightarrow \] \[P=100x-8{{x}^{3/2}}+C\] ?.(i) At initially \[x=0,\,\,\,\,P=1000\] Then, from Eq. (i) we get \[1000=100\times 0-0+C\] \[\Rightarrow \] \[C=1000\] On putting the value of C in Eq. (i), we get \[P=100x-8{{x}^{3/2}}+1000\] Now, at \[x=25,\] \[P=100(25)-8{{(25)}^{3/2}}+1000\] \[=2500-8(125)+100\] \[=3500-1000=2500\]You need to login to perform this action.
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