Quadratic equation word problem

In summary, the problem involves finding the rental price that will maximize income for Barkley's canoe-rental business. The business currently charges $12 per canoe and averages 36 rentals per day. For every $0.50 increase in rental price, the business expects to lose two rentals per day. The demand and income can be expressed in equations and the price that maximizes income can be found by solving for the maximum value of the income equation.
  • #1
musicgold
304
19
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.
 
Physics news on Phys.org
  • #2
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?
 
  • Like
Likes 1 person
  • #3
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]
 
  • Like
Likes 1 person
  • #4
Thank you.
 

Related to Quadratic equation word problem

1. What is a quadratic equation word problem?

A quadratic equation word problem is a type of math problem that involves using a quadratic equation (an equation with a variable raised to the power of two) to solve a real-life situation. These problems often involve finding the maximum or minimum value of a quantity, determining the height or distance of an object, or finding the roots or solutions to a problem.

2. How do I solve a quadratic equation word problem?

To solve a quadratic equation word problem, you need to first identify the unknown variable and assign it a variable (such as x). Then, write out the given information in terms of this variable. Next, use the quadratic formula (or other methods such as factoring or completing the square) to solve for the variable. Finally, check your answer to make sure it makes sense in the context of the problem.

3. What are the key components of a quadratic equation word problem?

The key components of a quadratic equation word problem are the unknown variable, the given information, and the quadratic equation itself. The unknown variable is typically the quantity that you are trying to solve for, while the given information provides context and necessary data for solving the problem. The quadratic equation is the mathematical expression that relates the unknown variable to the given information.

4. What are some real-life applications of quadratic equation word problems?

Quadratic equation word problems have many real-life applications, such as in physics (e.g. calculating the trajectory of a projectile), engineering (e.g. determining the optimal shape for a bridge), and finance (e.g. finding the maximum profit for a business). They can also be used to model natural phenomena, such as the growth of a population or the spread of a disease.

5. Are there any tips for solving quadratic equation word problems?

Yes, there are a few tips that can help you solve quadratic equation word problems more efficiently. First, make sure to carefully read the problem and identify the key components. Next, choose the most appropriate method for solving the problem (e.g. quadratic formula, factoring, completing the square). Additionally, it can be helpful to draw a diagram or create a table to organize the given information. Finally, always check your answer and make sure it makes sense in the context of the problem.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
20
Views
1K
Replies
4
Views
592
  • Precalculus Mathematics Homework Help
Replies
18
Views
2K
  • Precalculus Mathematics Homework Help
Replies
8
Views
821
  • Precalculus Mathematics Homework Help
Replies
19
Views
2K
  • Precalculus Mathematics Homework Help
Replies
6
Views
2K
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
4
Views
15K
  • Linear and Abstract Algebra
Replies
4
Views
2K
  • Precalculus Mathematics Homework Help
Replies
5
Views
2K
Back
Top