Revenue: price times quantity

Finance and economics, grade 12 (3 units) Lesson 1 of 16Learning

Revenue: price times quantity

A price rise that drives customers away

A business’s revenue is the money it takes from its sales: the price of one unit times the number of units sold.

A gym has 400 members, and each pays 100 ₪ a month.
The manager finds that every rise of 5 ₪ in the monthly price drives 10 members away.

$\displaystyle x$ is the number of 5 ₪ rises:

Rises $\displaystyle x$	0	2	4	6	8	10	12
Price (₪)	100	110	120	130	140	150	160
Members	400	380	360	340	320	300	280
Revenue (₪)	40,000	41,800	43,200	44,200	44,800	45,000	44,800

At first every rise increases the revenue: the higher price makes up for the members who left.
The revenue grows less and less, peaks at 10 rises, and then falls.

The revenue as a parabola

The revenue is the price times the quantity:

\displaystyle R=p\cdot q

$\displaystyle p$ – the price of one unit, $\displaystyle q$ – the number of units sold (here: members).

After $\displaystyle x$ rises the price is $\displaystyle 100+5x$ , and the number of members $\displaystyle 400-10x$ . So the revenue is

R(x)=(100+5x)(400-10x)

We expand the brackets:

R(x)=40000-1000x+2000x-50x^{2}=-50x^{2}+1000x+40000

$\displaystyle R(x)$ is a quadratic function: its graph is a parabola.
The coefficient of $\displaystyle x^{2}$ is negative, $\displaystyle {a}={-50}$ : the parabola has a highest point – the greatest revenue.
The coefficient is negative because the price goes up while the number of members goes down: the product $\displaystyle 5x\cdot(-10x)$ is a positive number times a negative one.

The graph of the revenue

On the graph, $\displaystyle x$ is the number of rises, and $\displaystyle y$ the revenue in shekels.

\displaystyle R(x)=(100+5x)(400-10x)

\displaystyle V=(10,45000)

The point $\displaystyle V$ is the parabola’s vertex: at 10 rises, a price of 150 ₪, the revenue is 45,000 ₪ – the greatest.
To the left of the vertex the revenue rises, and to its right it falls.
The parabola is symmetric about its vertex: at 8 rises and at 12 rises the revenue is the same, 44,800 ₪.
Only the part where the price and the number of members are positive describes the gym: $\displaystyle x$ between $\displaystyle -20$ and 40.

A quantity that falls as the price rises

Check: at 10 ₪,

\displaystyle {10\cdot300}={3000}

Sometimes you know directly how many units sell at each price.
A bakery sells $\displaystyle {q}={500-20p}$ loaves of bread a day, where $\displaystyle p$ is the price of a loaf in shekels.

Every extra shekel on the price takes 20 loaves a day off the sales: the quantity is a linear function of the price, decreasing.
The revenue is $\displaystyle {R(p)}={p(500-20p)}={500p-20p^{2}}$ – again a parabola opening downwards.

Price $\displaystyle p$ (₪)	5	10	15	20	25
Loaves $\displaystyle q$	400	300	200	100	0
Revenue (₪)	2,000	3,000	3,000	2,000	0

At 10 ₪ and at 15 ₪ the revenue is the same, so the vertex is halfway between them: at a price of 12.5 ₪ the revenue is $\displaystyle {12.5\cdot250}={3125}$ ₪.

Exercise 1

A cinema sells 300 tickets a screening, each for 40 ₪.
Every rise of 2 ₪ in the ticket price takes 10 tickets off the sales.
$\displaystyle x$ is the number of 2 ₪ rises.

What will the revenue from a screening be after 3 rises?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Write on the graph board the revenue from a screening after $\displaystyle x$ rises, $\displaystyle {y}={\dots}$

Is it worth the cinema’s while to raise the price by 2 ₪, once?
Explain with a calculation.

Exercise 2

A print shop sells 150 posters a week, each for 20 ₪.
Every extra shekel on the price takes 5 posters a week off the sales.
After $\displaystyle x$ rises of a shekel the revenue is $\displaystyle {R(x)}={(20+x)(150-5x)}$ .

Expand the brackets.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Which way does the parabola of the revenue open, up or down?
What does that say about the print shop’s revenue?

Exercise 3

A car wash washes $\displaystyle {q}={240-6p}$ cars a week, where $\displaystyle p$ is the price of a wash in shekels.

What is the weekly revenue when a wash costs 20 ₪?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

What is the weekly revenue when a wash costs 30 ₪?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

On the graph, $\displaystyle x$ is the price of a wash, and $\displaystyle y$ the weekly revenue.
Mark the points where the revenue is 0.

\displaystyle R(x)=x(240-6x)

What does each of the points you marked say about the car wash?

Formulas: finance and economics, grade 12 (3 units)

Open as a page

Revenue and profit: a quadratic model

The revenue – $\displaystyle p$ the price of a unit, $\displaystyle q$ the number of units sold
$\displaystyle R=p\cdot q$
The profit – the revenue minus the cost
$\displaystyle P=R-C$
A parabola – the revenue or the profit as a function of the price
$\displaystyle y=ax^{2}+bx+c$
The vertex’s $\displaystyle x$ – where the revenue or the profit is greatest (when $\displaystyle a<0$ )
$\displaystyle x=-\frac{b}{2a}$ $\displaystyle a\neq0$
The quadratic formula – the prices where the revenue or the profit is 0
$\displaystyle x_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

Linear programming

A constraint – an inequality; the points that satisfy it are a half-plane
$\displaystyle ax+by\le c$
The objective function – the greatest profit is at one of the feasible region’s corner points
$\displaystyle P=ax+by$

Analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$
The midpoint of the segment between the two points
$\displaystyle M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
The slope of the line through the two points
$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$
The equation of the line through the point $\displaystyle (x_1,y_1)$ with slope $\displaystyle m$
$\displaystyle y-y_1=m(x-x_1)$
Parallel lines – the same slope
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$

The normal distribution

In a normal distribution – $\displaystyle \mu$ the mean, $\displaystyle \sigma$ the standard deviation: about 68% of the data within one standard deviation of the mean (34% on each side)
$\displaystyle P(\mu-\sigma<x<\mu+\sigma)\approx68\%$
About 95% within two standard deviations of the mean (another 13.5% on each side)
$\displaystyle P(\mu-2\sigma<x<\mu+2\sigma)\approx95\%$
About 2.5% above $\displaystyle \mu+2\sigma$ , and about 2.5% below $\displaystyle \mu-2\sigma$
$\displaystyle P(x>\mu+2\sigma)\approx2.5\%$
The standard score of the value $\displaystyle x$ – how many standard deviations it is above the mean (or below it, when $\displaystyle z<0$ )
$\displaystyle z=\frac{x-\mu}{\sigma}$
The value whose standard score is $\displaystyle z$
$\displaystyle x=\mu+z\cdot\sigma$