Given
The demand function
[tex]p(x)=\frac{250}{\sqrt{x}}-5[/tex]
To determine:
a) The revenue function.
b) The marginal revenue.
c) The marginal revenue when x=200.
d) The equation of tangent, and its derivation.
Explanation:
It is given that,
[tex]p(x)=\frac{250}{\sqrt{x}}-5[/tex]
a) The revenue function is given by,
[tex]\begin{gathered} R(x)=x\cdot p(x) \\ R(x)=x\times(\frac{250}{\sqrt{x}}-5) \\ R(x)=250\sqrt{x}-5x \end{gathered}[/tex]
b) The marginal revenue function is,
[tex]\begin{gathered} R^{\prime}(x)=\frac{d}{dx}(R(x)) \\ =\frac{d}{dx}(250\sqrt{x}-5x) \\ =250\times(\frac{1}{2}\times x^{-\frac{1}{2}})-5 \\ =\frac{125}{\sqrt{x}}-5 \end{gathered}[/tex]
c) The marginal revenue when x=200 is,
[tex]\begin{gathered} R^{\prime}(200)=\frac{125}{\sqrt{200}}-5 \\ =\frac{125}{10\sqrt{2}}-5 \\ =8.8388-5 \\ =3.8388 \\ =3.84 \end{gathered}[/tex]
Hence, the marginal revenue is 3.84.
d) Let y=mx+c is the tangent.
Then,
[tex]y=3.84x+c[/tex]
That implies, for y=12.68, and x=200,
[tex]\begin{gathered} 12.68=3.84\times200+c \\ 12.68=768+c \\ c=12.68-768 \\ c=-755.32 \end{gathered}[/tex]
Hence, the equation of tangent is,
[tex]y=3.84x-755.32[/tex]
