Creating linear models

Use the information provided to answer Part A and Part B.

The Water Watch program is encouraging customers to reduce the amount of water they use each day. The program is selling low-flow showerheads, which use 2 gallons of water per minute, for $54 each.  

A family currently has a showerhead that uses 5 gallons of water per minute and is considering replacing it with one of the low-flow showerheads.  The family uses the shower an average of 20 minutes per day and pays $0.002 per gallon of water.

Part A
Create a model that can be used to determine the cost savings, in dollars, for the family to purchase and use a low-flow showerhead in terms of the number of days.

Then determine the number of days at which the family will start saving money.  Justify your answer in terms of the context.

Part B
One year after the low-flow showerhead is purchased, the cost of water increases by 5%. Create a new model to determine the cost savings, in dollars, with the increase in the cost of water.  

Use your model to determine the number of days at which the family will start saving money after the increase in the cost of water.  Justify your answer.

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Part A
For 20 minutes of shower time,
the family can save:
(5 – 2)(20)   {five gallons minus 2 gallons}
= 60 gallons a day
and
at $0.002 per gallon, that is
60(0.002) in savings
= $0.12 savings per day

S = savings in dollars
d = number of days

S = 0.12d – 54  {model is savings minus cost}
0 = 0.12d – 54  {savings is 0 for break-even}
54 = 0.12d   {added 54 to each side}
d = 450   {divided each side by 0.12}


In 450 days, the amount saved will
be equal to the cost of the low-flow showerhead.
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Ask Algebra House

Part B

Savings for 1st year
$365(0.12)   {saving $0.12 per day}
= $43.80 saved in 1st year

The showerhead is $54, they still have 54 – 43.8, or $10.20 to make up.

After 1st year
12(1.05)   {water increases by 5%}
= $0.126 per day saved

S = savings
y = number of days in year 2

Number of days at which savings becomes zero
0.126y – 10.2 = S  {new model}
0.126y – 10.2 = 0   {set savings to 0, break-even}
0.126y = 10.2   {added 10.2 to each side}
y ≈ 80.96 days   {divided each side by 0.126}

Family began saving 81 days into second year.
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