smc-5237-40-Cost/Revenue

v1 Algebra, STD2 A4 2010 HSC 24b

Damo hires paddle boats in summertime as part of his water sports business. To calculate the cost, $C$, in dollars, of hiring $x$ paddle boats, he uses the equation $C=40+25x$.

He hires the paddle boats for $35 per hour and determines his income, $I$, in dollars, using the equation $I=35x$.

Use the graph to solve the two equations simultaneously for $x$ and explain the significance of this solution for Damo's business. (2 marks)

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$x=4\ \ \text{See worked solution}$

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$\text{From the graph, intersection occurs at}\ x=4$

$\rightarrow\ \text{Break-even point occurs at}\ x=4$

$\text{i.e. when 4 hours of paddle board hire occurs}$

$\text{Income}$	$=35\times 4=$140\ \ \text{is equal to}$
$\text{Costs}$	$=40+(25\times 4)=$140$

$\text{If}\ <4\ \text{hours of board hire}\ \rightarrow\ \text{LOSS for business}$

$\text{If}\ >4\ \text{hours of board hire}\ \rightarrow\ \text{PROFIT}$

♦ Mean mark 36%.
MARKER’S COMMENT: The intersection on the graph is the same point at which the two simultaneous equations are solved for the given value of $x$.

v1 Algebra, STD2 A4 SM-Bank 4

Bec is a baker and makes cookies to sell every week.

The cost of making $n$ cookies, $$C$, can be calculated using the equation

$C=400+2.5n$

Bec sells the cookies for $4.50 each, and her income is calculated using the equation

$I=450n$

On the grid above, draw the graphs of $C$ and $I$. (2 marks)

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On the graph, label the breakeven point and the loss zone. (2 marks)

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(i) and (ii)

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ii. $\text{Loss zone occurs when}\ C > I,\ \text{which is shaded}$

$\text{in the diagram above.}$

v1 Algebra, STD2 A4 2005 HSC 28b

Jake and Preston are planning a fund-raising event at the local swim centre. They can have access to the giant pool float for $550 and the party room hire for $250. A sausage sizzle and drinks will cost them $9 per person.

Write a formula for the cost ($C) of running the event for $x$ people. (1 mark)

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The graph shows planned income and costs when the ticket price is $15.

Estimate the minimum number of people needed at the fund raising event to cover the costs. (1 mark)

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How much profit will be made if 200 people attend the fund raiser? (1 mark)

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Jake and Preston have 300 tickets to sell. They want to make a profit of $1510.

What should be the price of a ticket, assuming all 300 tickets will be sold? (3 marks)

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$C=800+9x$
$\text{Approximately }135$
$$400$
$$16.70$

Show Worked Solution

i.	$$C$	$=550+250+(9\times x)$
		$=800+9x$

ii. $\text{Using the graph intersection}$

$\text{Approximately 135 people are needed}$

$\text{to cover the costs.}$

iii. $\text{If 200 people attend}$

$\text{Income}$	$=200\times $15$
	$=$3000$

$\text{Costs}$	$=800+(9\times 200)$
	$=$2600$

$\therefore\ \text{Profit}$	$=3000-2600$
	$=$400$

iv. $\text{Costs when}\ x=300:$

$C$	$=800+(9\times 300)$
	$=$3500$

$\text{Income required to make }$1510\ \text{profit}$

$=3500+1510$

$=$5010$

$\therefore\ \text{Price per ticket}$	$=\dfrac{5010}{300}$
	$=$16.70$

v1 Algebra, STD2 A4 2019 HSC 36

A small business makes and dog kennels.

Technology was used to draw straight-line graphs to represent the cost of making the dog kennels $(C)$ and the revenue from selling dog kennels $(R)$. The $x$-axis displays the number of dog kennels and the $y$-axis displays the cost/revenue in dollars.

How many dog kennels need to sold to break even? (1 mark)

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By first forming equations for cost `(C)` and revenue `(R)`, determine how many dog kennels need to be sold to earn a profit of $2500. (3 marks)

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$20$
$145$

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a. $20\ \ (x\text{-value at intersection})$

b. $\text{Find equations of both lines}:$

$(0, 400)\ \text{and}\ (20, 600)\ \text{lie on}\ \ C$

$\text{gradient}_C = \dfrac{600-400}{20-0}=10$

$\rightarrow\ C=400+10x$

$(0,0)\ \text{and}\ (20, 600)\ \text{lie on}\ \ R$

$\text{gradient}_R =\dfrac{600-0}{20-0}=30$

$\rightarrow\ R=30x$

$\text{Profit} = R-C$

$\text{Find}\ \ x\ \text{when Profit }= $2500:$

$2500$	$=30x-(400+10x)$
$20x$	$=2900$
$x$	$=145$

$\therefore\ 145\ \text{dog kennels need to be sold to earn }$2500\ \text{profit}$

♦♦ Mean mark 28%.

v1 Algebra, STD2 A4 2020 VCAA 3

Noah's business manufactures car seat covers.

The monthly oncome, $I$, in dollars, from selling $n$ seat covers is given by

$I=60n$

This relationship is shown on the graph below.

The monthly cost, $C$, in dollars, of making $n$ seat covers is given by

$C=35n+5000$

On the graph above, sketch the monthly cost, $C$, of making $n$ seat covers. (1 mark)
Find the number of seat covers that need to be sold in order to break even and state the profit made at this point. (2 marks)

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b. $\text{200 seat covers and zero profit at this point}$

Show Worked Solution

a. $\text{Draw graph through points (0, 5000) and (250, 13 750)}$

b. $C=35n+5000\ \text{and }I=60n$

$\text{Break-even occurs when} \ \ I=C$

$\text{Method 2: Graphically}$

$\text{Point of intersection is }\rightarrow (200, 12\ 000)$

$\text{Method 2: Algebraically}$

$\text{Solve for} \ n:$

$60n$	$=35n+5000$
$25n$	$=5000$
$n$	$=200$

$\therefore\ \text{200 seat covers must be sold to break even}$

$\text{and the profit at this point is zero}$

v1 Algebra, STD2 A4 SM-Bank 27

Morgan and Beau are to host a 21st birthday party for their friend's Zac and Peattie. They can hire a function room for $900 and a DJ for $450. Drinks will cost them $33 per person.

Write a formula for the cost ($$C$) of holding the birthday party for $x$ people. (1 mark)

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The graph below shows the planned income and costs if they charge $60 per person. Estimate the number of friends they need to invite to break even. (1 mark)

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How much money will Morgan and Beau have to purchase a travel voucher as a group gift if 90 people attend the party? (1 mark)

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$$C=1350+33x$
$50$
$$1080$

Show Worked Solution

i.	$\text{Fixed Costs}$	$=900+450$
		$=$1350$

$\text{Variable Costs}=$33x$

$\therefore\ $C=1350+33x$

ii.	$\text{From the graph}$
	$\text{Costs = Income when}\ x=50$
	$\text{(i.e. where lines intersect)}$

$\text{Algebraically}$
$\text{Income }$	$=\ \text{Costs}$
$60x$	$=1350+33x$
$27x$	$=1350$
$x$	$=50$

$\therefore\ \text{Breakeven when }50\ \text{people attend}$

iii. $\text{When}\ \ x=90:$

$\text{Income}$	$=90\times 60$
	$=$5400$

$$C$	$=1350+33\times 90$
	$=$4320$

$\therefore\ \text{Travel Voucher}$	$=5400-4320$
	$=$1080$

v1 Algebra, STD2 A4 2018 HSC 27d

The graph displays the cost ($$c$) charged by two companies for the hire of a jetski for $x$ hours.

Both companies charge $450 for the hire of a jetski for 5 hours.

What is the hourly rate charged by Company A? (1 mark)

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Company B charges an initial booking fee of $80.

Write a formula, in the of $c=b+mx$, for the cost of hiring a jetski from Company B for $x$ hours. (2 marks)

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A jetski is hired for 7 hours from Company B.

Calculate how much cheaper this is than hiring from Company A. (2 marks)

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$$90$
$c=80+74x$
$$32$

Show Worked Solution

i.	$\text{Hourly rate}\ (A)$	$=\dfrac{450}{5}$
		$=$90$

ii. $m=\text{hourly rate}$

$\text{Find}\ m,\ \text{given}\ c = 450,\ \text{when}\ \ x = 5\ \text{and}\ \ b = 80$

$450$	$=80+m\times 5$
$5m$	$=370$
$m$	$=\dfrac{370}{5}=74$

$\therefore\ c=80+74x$

iii.	$\text{Cost}\ (A)$	$=90\times 7=$630$
	$\text{Cost}\ (B)$	$=80+74\times 7=$598$

$\therefore\ \text{Company}\ B’\text{s hiring cost is }$32\ \text{cheaper.}$

v1 Algebra, STD2 A4 2011 HSC 20 MC

A function centre hosts events for up to 500 people. The cost $C$, in dollars, for the centre
to host an event, where $x$ people attend, is given by:

$C=20\ 000+40x$

The centre charges $120 per person. Its income $I$, in dollars, is given by:

$I=120x$

How much greater is the income of the function centre when 500 people attend an event, than its income at the breakeven point?

$$10\ 000$
$$20\ 000$
$$30\ 000$
$$40\ 000$

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$C$

Show Worked Solution

$\text{When}\ x=500,\ I=120\times 500=$60\ 000$

$\text{Breakeven when}\ \ x=250\ \ \text{(from graph)}$

$\text{When}\ \ x=250,\ I=120\times 250=$30\ 000$

$\text{Difference}$	$=60\ 000-30\ 000$
	$=$30\ 000$

$\Rightarrow C$

♦ Mean mark 50%
COMMENT: Students can read the income levels directly off the graph to save time and then check with the equations given.

\(\text{Income}\)	\(=35\times 4=$140\ \ \text{is equal to}\)
\(\text{Costs}\)	\(=40+(25\times 4)=$140\)

i.	\($C\)	\(=550+250+(9\times x)\)
		\(=800+9x\)

\(2500\)	\(=30x-(400+10x)\)
\(20x\)	\(=2900\)
\(x\)	\(=145\)

\(60n\)	\(=35n+5000\)
\(25n\)	\(=5000\)
\(n\)	\(=200\)

i.	\(\text{Fixed Costs}\)	\(=900+450\)
		\(=$1350\)

\(\text{Income}\)	\(=200\times $15\)
	\(=$3000\)

\(\text{Costs}\)	\(=800+(9\times 200)\)
	\(=$2600\)

\(\therefore\ \text{Profit}\)	\(=3000-2600\)
	\(=$400\)

\(C\)	\(=800+(9\times 300)\)
	\(=$3500\)