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r-ruslan [8.4K]
1 year ago
5

Need help with courseReplace 250 with 575 and -3 with -5.

Mathematics
1 answer:
Svetlanka [38]1 year ago
8 0

Given

The demand function

p(x)=\frac{250}{\sqrt{x}}-5

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,

p(x)=\frac{250}{\sqrt{x}}-5

a) The revenue function is given by,

\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}

b) The marginal revenue function is,

\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}

c) The marginal revenue when x=200 is,

\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}

Hence, the marginal revenue is 3.84.

d) Let y=mx+c is the tangent.

Then,

y=3.84x+c

That implies, for y=12.68, and x=200,

\begin{gathered} 12.68=3.84\times200+c \\ 12.68=768+c \\ c=12.68-768 \\ c=-755.32 \end{gathered}

Hence, the equation of tangent is,

y=3.84x-755.32

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