MHT CET · Maths · Differentiation
At present a firm is manufacturing 1000 items. It is estimated that the rate of change of production P w.r.t. additional number of worker \(x\) is given by \(\frac{\mathrm{dp}}{\mathrm{d} x}=100-12 \sqrt{x}\).
If the firm employees 9 more workers, then the new level of production of items is
- A \(1684\)
- B \(1648\)
- C \(2116\)
- D \(1116\)
Answer & Solution
Correct Answer
(A) \(1684\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{dP}}{\mathrm{d} x}=100-12 \sqrt{x} \)
\( \text { Integrating both sides, we get } \)
\( \int \mathrm{dp}=\int(100-12 \sqrt{x}) \mathrm{d} x \)
\( \therefore \mathrm{P}=100 x-8 x \sqrt{x}+\mathrm{c} \)
\( \text { Given that } \mathrm{P}=1000, \text { when } x=0 \)
\( \Rightarrow 1000=100(0)-8(0)+\mathrm{c} \)
\( \Rightarrow \mathrm{c}=1000 \)
\( \therefore \mathrm{P}=100 x-8 x \sqrt{x}+1000 \)
\( \text { When } x=9, \text { we get } \)
\( \mathrm{P}=900-216+1000=1684\)
\(\therefore \) The new level of production of items is 1684 .
\( \text { Integrating both sides, we get } \)
\( \int \mathrm{dp}=\int(100-12 \sqrt{x}) \mathrm{d} x \)
\( \therefore \mathrm{P}=100 x-8 x \sqrt{x}+\mathrm{c} \)
\( \text { Given that } \mathrm{P}=1000, \text { when } x=0 \)
\( \Rightarrow 1000=100(0)-8(0)+\mathrm{c} \)
\( \Rightarrow \mathrm{c}=1000 \)
\( \therefore \mathrm{P}=100 x-8 x \sqrt{x}+1000 \)
\( \text { When } x=9, \text { we get } \)
\( \mathrm{P}=900-216+1000=1684\)
\(\therefore \) The new level of production of items is 1684 .
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