Quadratic equation word problem

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musicgold
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1. Problem
Barkley runs a canoe-rental business on a small river in Pennsylvania. Currently, the
business charges $12 per canoe and they average 36 rentals a day. A study shows that
for every $.50 increase in rental price, the business can expect to lose two rentals per
day. Find the price that will maximize income.

2. The attempt at a solution
I found this problem on the internet. I can probably solve the problem manually, but I want to learn how to create a set of equation to describe this situation.
R= revenue, x= rental rate and t= number of rentals

R = x * t
R= 432 in a normal state.

I am not sure how to link a $0.50 change in x to a 2 unit change in t.

How should I go about this?

Thanks.
 
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If you increase the price by $0.50 then your revenue will now be

[tex]R=(x+0.5)(t-2)[/tex]

Increase it again by another $0.50 and you now have?

What if you increase it n times?
 
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musicgold said:
1. Problem
Barkley runs a canoe-rental business on a small river in Pennsylvania. Currently, the
business charges $12 per canoe and they average 36 rentals a day. A study shows that
for every $.50 increase in rental price, the business can expect to lose two rentals per
day. Find the price that will maximize income.

2. The attempt at a solution
I found this problem on the internet. I can probably solve the problem manually, but I want to learn how to create a set of equation to describe this situation.
R= revenue, x= rental rate and t= number of rentals

R = x * t
R= 432 in a normal state.

I am not sure how to link a $0.50 change in x to a 2 unit change in t.

How should I go about this?

Thanks.

The demand (number of rentals per day) is given by
[tex]N = 36 - 4(p-12)[/tex]
where ##p =## price ($) and ##N=## number rented per day. Note that when ##p = 12## we have ##N = 36##, as given in the problem. Note also that ##N## decreases by 4 when ##p## increases by 1 (that is, ##N## decreases by 2 when ##p## increases by 0.5).

The income is ##I = p N## because we have ##N## rentals and receive ##p##($) for each rental. Thus the daily income (in $) is
[tex]I = pN = p[36 - 4(p-12)] = 84 p - 4 p^2[/tex]
 
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