question_answer

Find the equation of tangent to the curve give b by \[x=a{{\sin }^{3}}t\] and \[y=b{{\cos }^{3}}t\] at  a point \[t=\pi /2.\]

OR

A telephone company in a town has 500 subscribers on its list and collects fixed charge subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of Rs. 1, one subscriber will

Discontinue the service. Find what increase will bring maximum profit?

Answer:

Given, \[x=a{{\sin }^{3}}t\] and \[y=b{{\cos }^{3}}t\]    ?(i)

On differentiating both sides of Eq. (i) w.r.t. t, we get

\[\frac{dx}{dt}=3\,a\,{{\sin }^{2}}t\frac{d}{dt}(sin\,t)=3\,asi{{n}^{2}}t\,\cos t\]

and \[\frac{dy}{dt}=3\,b\,{{\cos }^{2}}t\frac{d}{dt}(cos\,t)=-\,3\,b{{\cos }^{2}}t\,\sin t\]

Now, \[\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=-\,\frac{3\,b\,{{\cos }^{2}}t\cdot \sin t}{3\,a\,{{\sin }^{2}}t\cdot \cos t}=-\,\frac{b}{a}\cot \,t\]

At \[t=\pi /2,\]

Slope of tangent, \[{{\left( \frac{dy}{dx} \right)}_{at\,\,\pi /2}}=-\,\frac{b}{a}\cot \frac{\pi }{2}=0\]

At \[t=\frac{\pi }{2},\] point of contact is \[\left( a{{\sin }^{3}}\frac{\pi }{2},\,\,b{{\cos }^{3}}\frac{\pi }{2} \right).\]

\[\therefore \] Required equation o tangent is\[y-b{{\cos }^{3}}\frac{\pi }{2}={{\left( \frac{dy}{dx} \right)}_{at\frac{\pi }{2}}}\left( x-a{{\sin }^{3}}\frac{\pi }{2} \right)\]

[\[\because \] equation of line is \[(y-{{y}_{1}})\]\[\left. ={{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}}(x-{{x}_{1}}) \right]\]

\[\Rightarrow \] \[(y-b\times 0)=0(x-a\times 1)\]

\[\left[ \because \,\,\,\cos \frac{\pi }{2}=0,\,\,\sin \frac{\pi }{2}=1 \right]\]

\[\Rightarrow \]   \[y-0=0\] \[\Rightarrow \] y = 0

OR

Let the company increases the annual subscription fee by Rs. x. Then, x subscribers will discontinue the service.

\[\therefore \] Total revenue of company after the increment is given by

\[R(x)=(500-x)(300+x)\]

\[=500\times 300+500x-300x-{{x}^{2}}\]

\[=15\times {{10}^{4}}+200x-{{x}^{2}}\]

\[=-{{x}^{2}}+200x+150000\]

On differentiating both sides w.r.t.x, we get

\[R'(x)=-\,2x+200\]

For maxima or minima, put R?(x) = 0

\[\Rightarrow \]   \[-\,2x+200=0\]

\[\Rightarrow \]   x = 100

Again, differentiating Eq. (i) w.r.t.x we get

\[R''(x)=-\,2\]

\[\Rightarrow \]   \[R''(100)=-\,2<0\]

So, R(x) is maximum when x =100.

Hence, the company should increase the subscription fee by Rs 100, so that it has maximum profit.