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3 years ago

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Pearl's Biking Company manufactures and sells bikes. Each bike costs £40 to make, and the company's fixed costs are £5000. In ad

dition, Pearl knows that the price of each bike comes from the price function P(x) = 300 – 2x where x is the total number of bikes manufactured and sold. Questions a) Find the company's revenue function R(x), the company's cost function C(x) and find the breakeven points

1 answer:

____ [38]3 years ago

3 0

Answer:

R(x) =300·x - 2·x²

C(x) = £5000 + £40 × x

The break even points are 23.47 and 106.53 or 23 and 107 bikes

Step-by-step explanation:

Given that the price function P(x) = 300 -2·x

Cost per bike = £40

The revenue function R(x) is given by bike price × total number of bikes manufactured and sold

∴ R(x) = P(x)×x = (300 - 2·x)×x = 300·x - 2·x²

The company's cost function, C(x) is Fixed cost + cost to produce each bike × total number of bikes produced

∴ C(x) = £5000 + £40 × x

The break even point is given by the relation;

Total revenue - total cost = 0

That is, break even point is R(x) - C(x) = 0

300·x - 2·x² - (5000 + 40·x) = 0

-2·x²+260·x-5000 = 0 or 2·x²- 260·x + 5000 = 0

Factorizing, we have;

(x - (65 -5√69))(x - (65 +5√69))

Solving gives x = 23.47 or 106.53

Therefore, the break even points are 23.47 and 106.53.

That is the company is profitable when they produce less than 23 bikes or more than 107 bikes.

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2 years ago

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A coin is flipped 10 times where each flip comes up either heads or tails. How many possible outcomes (a) contain exactly two he

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Answer:

a. 45

b. 176

c. 252

Step-by-step explanation:

First take into account the concept of combination and permutation:

In the permutation the order is important and it is signed as follows:

P (n, r) = n! / (n - r)!

In the combination the order is NOT important and is signed as follows:

C (n, r) = n! / r! (n - r)!

Now, to start with part a, which corresponds to a combination because the order here is not important. Thus

n = 10

r = 2

C (10, 2) = 10! / 2! * (10-2)! = 10! / (2! * 8!) = 45

There are 45 possible scenarios.

Part b, would also be a combination, defined as follows

n = 10

r <= 3

Therefore, several cases must be made:

C (10, 0) = 10! / 0! * (10-0)! = 10! / (0! * 10!) = 1

C (10, 1) = 10! / 1! * (10-1)! = 10! / (1! * 9!) = 10

C (10, 2) = 10! / 2! * (10-2)! = 10! / (2! * 8!) = 45

C (10, 3) = 10! / 3! * (10-3)! = 10! / (2! * 7!) = 120

The sum of all these scenarios would give us the number of possible total scenarios:

1 + 10 + 45 + 120 = 176 possible total scenarios.

part c, also corresponds to a combination, and to be equal it must be divided by two since the coin is thrown 10 times, it would be 10/2 = 5, that is our r = 5

Knowing this, the combination formula is applied:

C (10, 5) = 10! / 5! * (10-5)! = 10! / (2! * 5!) = 252

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3 years ago

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Answer:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

Step-by-step explanation:

Given

$f(x)= \frac{9}{3x+ 2}$

$c = 6$

The geometric series centered at c is of the form:

$\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.$

Where:

$a \to$ first term

$r - c \to$ common ratio

We have to write

$f(x)= \frac{9}{3x+ 2}$

In the following form:

$\frac{a}{1 - r}$

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$f(x)= \frac{9}{3x+ 2}$

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$f(x) = \frac{9}{3x - 18 + 18 +2}$

$f(x) = \frac{9}{3x - 18 + 20}$

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$f(x) = \frac{1}{\frac{1}{9}(3x + 2)}$

Open bracket

$f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}$

Rewrite as:

$f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}$

Collect like terms

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}$

Take LCM

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}$

$f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}$

So, we have:

$f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}$

By comparison with: $\frac{a}{1 - r}$

$a = 1$

$r = -\frac{1}{3}x + \frac{7}{9}$

$r = -\frac{1}{3}(x - \frac{7}{3})$

At c = 6, we have:

$r = -\frac{1}{3}(x - \frac{7}{3}+6-6)$

Take LCM

$r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)$

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)

So, the power series becomes:

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}ar^n$

Substitute 1 for a

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}1*r^n$

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}r^n$

Substitute the expression for r

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n$

Expand

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]$

Further expand:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................$

The power series converges when:

$\frac{1}{3}|x - \frac{7}{3}| < 1$

Multiply both sides by 3

$|x - \frac{7}{3}|$

Expand the absolute inequality

$-3 < x - \frac{7}{3}$

Solve for x

$\frac{7}{3} -3 < x$

Take LCM

$\frac{7-9}{3} < x$

$-\frac{2}{3} < x$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

6 0

2 years ago