QUIZ 1 The market research department of a company recommended to management that the
company manufacture and market a promising product on the market. After extensive surveys, the
research departmeni backed up the recommendation with the demand equation;
x = f(p) = 6,000 – 30p
where x is the number of units that retailers are likely to buy per month at GHSp per unis. From the
finance department, the following cost equation was obtained:
C = g(x) = 72,000 + 60%
Assuming revenue equation is R = xp
(a) Find the profit equation equations in terms of p only.
(5 marks)
(b) Calculate the price (s) and the quantity demanded at which the company would break oven.
(6 marks)
(c) Ar what price does the company maximize prot?
15 marks)

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Ответ:

kylucienne

08.04.2020

Mathematics

$\mathsf{ \blue{ f'(x) = \frac{ - 10}{ {(x - 5)}^{2} } } }$

Step-by-step explanation:

$\mathsf{f(x) = \frac{ {x}^{2} + 10x + 25}{ {x}^{2} - 25} }$

the above expression can be reduced to simpler terms

x² - 25 = (x + 5)(x - 5)x² + 10x + 25 = (x + 5)²

$\mathsf{\implies f(x) = \frac{ {(x + 5)}^{2} }{(x + 5)(x - 5)} }$

(x + 5)² can be written as (x + 5)(x + 5)

$\mathsf{\implies f(x) = \frac{ \cancel{(x + 5)}(x + 5)}{\cancel{(x + 5)}(x - 5)} }$

$\mathsf{\implies f(x) = \frac{x + 5}{x - 5} }$

Derivative of a fraction $\mathsf{\frac{u}{v}}$ is

$\boxed { \red {\mathsf{ \frac{v(\frac{du}{dx}) \: - \: u(\frac{dv}{dx})}{v^2}} }}$

$\mathsf{\implies f'(x) = \frac{(x - 5) - (x + 5)}{(x - 5) {}^{2} } }$

$\mathsf{\implies f '(x) = \frac{x - 5 - x - 5}{ {(x - 5)}^{2} } }$

$\mathsf{\implies f'(x) = \frac{ - 10}{ {(x - 5)}^{2} } }$

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