Quadratic equation word problem

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Homework Help Overview

The problem involves maximizing income for a canoe-rental business based on price adjustments and rental rates. It is situated within the context of quadratic equations and revenue calculations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss how to formulate equations that relate rental price changes to rental quantity changes. There is an exploration of revenue calculations based on price and demand adjustments.

Discussion Status

Some participants have provided insights into how to express revenue as a function of price and rentals, while others are questioning how to effectively link price increases to changes in rental numbers. Multiple interpretations of the problem are being explored.

Contextual Notes

The original poster expresses uncertainty about connecting price changes to rental changes, and there is a focus on deriving a quadratic equation from the given scenario. The problem is framed within the constraints of a real-world business model.

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

R=(x+0.5)(t-2)

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
N = 36 - 4(p-12)
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
I = pN = p[36 - 4(p-12)] = 84 p - 4 p^2
 
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Thank you.
 

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