Respuesta :
Consider that the revenue is calculated as,
[tex]\text{Revenue}=\text{Price per student}\times\text{No. of student}[/tex]The current revenue of Mrs. Nicely is,
[tex]R_{\circ}=20\times21=420[/tex]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,
[tex]R(x)=(20+4x)(21-x)=-4x^2+64x+420[/tex]Now, obtain the stationary points as,
[tex]\begin{gathered} R^{\prime}(x)=0 \\ -8x+64=0 \\ 8x=64 \\ x=8 \end{gathered}[/tex]Apply the second derivative check,
[tex]\begin{gathered} R^{\doubleprime}(x)=\frac{d \square}{dx}(R^{\prime}(x)) \\ R^{\doubleprime}(x)=\frac{d\square}{dx}(-8x+64) \\ R^{\doubleprime}(x)=-8 \\ R^{\doubleprime}(x)<0 \\ R^{\doubleprime}(8)<0 \end{gathered}[/tex]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,
[tex]P=20+4x=20+4(8)=20+32=52[/tex]So Mrs. Nicely should charge $52 for tutoring in order to maximize her revenue.
And the maximum revenue is calculated as,
[tex]R_{\max }(x)=R(8)=52\times13=676[/tex]Thus, the maximum revenue she can expect is $676 each week.
