JEE MAIN - Mathematics (2013 (Offline) - No. 4)
At present, a firm is manufacturing $$2000$$ items. It is estimated that the rate of change of production P w.r.t. additional number of workers $$x$$ is given by $${{dp} \over {dx}} = 100 - 12\sqrt x .$$ If the firm employs $$25$$ more workers, then the new level of production of items is
$$2500$$
$$3000$$
$$3500$$
$$4500$$
Explanation
Given, Rate of change is $${{dp} \over {dx}} = 100 - 12\sqrt x $$
$$ \Rightarrow dP = \left( {100 - 12\sqrt x } \right)dx$$
By intergrating $$\int {dP = \int {\left( {100 - 12\sqrt x } \right)} } dx$$
$$\int {dP} = \int {\left( {100 - 12\sqrt x } \right)} dx$$
$$P = 100x - 8{x^{3/2}} + C$$
Given, when $$x=0$$ then $$P=2000$$
$$ \Rightarrow C = 2000$$
Now when $$x$$$$=25$$
then $$P = 100 \times 25 - 8 \times {\left( {25} \right)^{3/2}} + 2000$$
$$\,\,\,\,\,\,\,\,\,\,$$ $$=4500-1000$$
$$ \Rightarrow P = 3500$$
$$ \Rightarrow dP = \left( {100 - 12\sqrt x } \right)dx$$
By intergrating $$\int {dP = \int {\left( {100 - 12\sqrt x } \right)} } dx$$
$$\int {dP} = \int {\left( {100 - 12\sqrt x } \right)} dx$$
$$P = 100x - 8{x^{3/2}} + C$$
Given, when $$x=0$$ then $$P=2000$$
$$ \Rightarrow C = 2000$$
Now when $$x$$$$=25$$
then $$P = 100 \times 25 - 8 \times {\left( {25} \right)^{3/2}} + 2000$$
$$\,\,\,\,\,\,\,\,\,\,$$ $$=4500-1000$$
$$ \Rightarrow P = 3500$$
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