Question

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 $$\frac{{dP}}{{dx}} = 100 - 12\sqrt x .$$     If the firm employs 25 more workers, then the new level of production of items is-

A. $$2500$$
B. $$3000$$
C. $$3500$$  
D. $$4500$$
Answer :   $$3500$$
Solution :
Given, Rate of change is
$$\eqalign{ & \frac{{dP}}{{dx}} = 100 - 12\sqrt x \cr & \Rightarrow dP = \left( {100 - 12\sqrt x } \right)dx \cr} $$
By integrating
$$\eqalign{ & \Rightarrow \int {dP = \int {\left( {100 - 12\sqrt x } \right)dx} } \cr & P = 100x - 8{x^{\frac{3}{2}}} + C \cr} $$
Given, when $$x=0$$   then $$P =2000$$
$$ \Rightarrow C = 2000$$
Now when $$x= 25$$  then
$$\eqalign{ & P = 100 \times 25 - 8 \times {\left( {25} \right)^{\frac{3}{2}}} + 2000 \cr & \Rightarrow P = 4500 - 1000 \cr & \Rightarrow P = 3500 \cr} $$

Releted MCQ Question on
Calculus >> Differential Equations

Releted Question 1

A solution of the differential equation $${\left( {\frac{{dy}}{{dx}}} \right)^2} - x\frac{{dy}}{{dx}} + y = 0$$     is-

A. $$y=2$$
B. $$y=2x$$
C. $$y=2x-4$$
D. $$y = 2{x^2} - 4$$
Releted Question 2

If $${x^2} + {y^2} = 1,$$   then

A. $$yy'' - 2{\left( {y'} \right)^2} + 1 = 0$$
B. $$yy'' + {\left( {y'} \right)^2} + 1 = 0$$
C. $$yy'' + {\left( {y'} \right)^2} - 1 = 0$$
D. $$yy'' + 2{\left( {y'} \right)^2} + 1 = 0$$
Releted Question 3

If $$y\left( t \right)$$ is a solution $$\left( {1 + t} \right)\frac{{dy}}{{dt}} - ty = 1$$    and $$y\left( 0 \right) = - 1,$$   then $$y\left( 1 \right)$$ is equal to-

A. $$ - \frac{1}{2}$$
B. $$e + \frac{1}{2}$$
C. $$e - \frac{1}{2}$$
D. $$\frac{1}{2}$$
Releted Question 4

If $$y = y\left( x \right)$$   and $$\frac{{2 + \sin \,x}}{{y + 1}}\left( {\frac{{dy}}{{dx}}} \right) = - \cos \,x,\,y\left( 0 \right) = 1,$$
then $$y\left( {\frac{\pi }{2}} \right)$$   equals-

A. $$\frac{1}{3}$$
B. $$\frac{2}{3}$$
C. $$ - \frac{1}{3}$$
D. $$1$$

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Differential Equations


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