Question

mrs. nicely tutors for $20 an hour and is currently tutoring 21 students a week. she decides that if she raises her prices she can tutor less students and maximize her profits. from her own observations, for every $4 increase in price she will lose 1 student per week. what price should mrs.nicely charge for tutoring to maximize her revenue? what's the maximum amount of revenue she can expect each week?

Answer 1

Consider that the revenue is calculated as,

`\text{Revenue}=\text{Price per student}*\text{No. of student}`

The current revenue of Mrs. Nicely is,

`R_(\circ)=20*21=420`

Let Mrs. Nicely lost 'x' students due to the fee hike. Now, there are (21-x) students remaining in her.

Given that every $4 increase in price results in loss of 1 student. The initial price was $20. So the price after 'x' students left will be (20+4x)$.

So the revenue function becomes,

`R(x)=(20+4x)(21-x)=-4x^2+64x+420`

Now, obtain the stationary points as,

`\begin{gathered} R^(\prime)(x)=0 \\ -8x+64=0 \\ 8x=64 \\ x=8 \end{gathered}`

Apply the second derivative check,

`\begin{gathered} R^(\doubleprime)(x)=(d \square)/(dx)(R^(\prime)(x)) \\ R^(\doubleprime)(x)=(d\square)/(dx)(-8x+64) \\ R^(\doubleprime)(x)=-8 \\ R^(\doubleprime)(x)<0 \\ R^(\doubleprime)(8)<0 \end{gathered}`

Since the second derivative is less than zero, the revenue function is maximum at x=8.

Thus, Mrs Nicely will have maximum revenue when 8 students have left, and the remaining students are (21-8=)13.

The corresponding price per student is calculated as,

`P=20+4x=20+4(8)=20+32=52`

So Mrs. Nicely should charge $52 for tutoring in order to maximize her revenue.

And the maximum revenue is calculated as,

`R_(\max )(x)=R(8)=52*13=676`

Thus, the maximum revenue she can expect is $676 each week.