My Math Forum  

Go Back   My Math Forum > High School Math Forum > Elementary Math

Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

LinkBack Thread Tools Display Modes
April 13th, 2018, 06:31 PM   #1
Chemist116's Avatar
Joined: Jun 2017
From: Lima, Peru

Posts: 40
Thanks: 1

Math Focus: Calculus
Question How to find the least number of objects in a problem involving profits?

The problem is as follows:

In an electronics factory, the owner calculates that the cost to produce his new model of portable TV is $26$ dollars. After meeting with the distributors, he agrees the sale price for his new product to be $25$ dollars each and additionally $8%$ more for each TV set sold after $8000$ units. What is the least number of TV's he has to sell in order to make a profit?.

The answers are:
  • 16000
  • 15001
  • 16001
  • 15999
  • 17121

This problem has made me to go in circles on how to express it in a mathematical expression. I'm not sure if it does need to use of inequations.

What I tried to far is to think this way:

The first scenario is what if what he sells is $8000$ units, then this would become into:

$$\textrm{production cost:}\,26\frac{\$}{\textrm{unit}} \times 8000\,\textrm{units} = 208000 \$$$

$$\textrm{sales:}\,25\frac{\$}{\textrm{unit}} \times 8000\,\textrm{units}= 200000\$$$

Therefore there will be an offset of $8000\,\$$ as


So I thought what If I consider the second part of the problem which it says that he will receive an additional of 8% after $8000$ units.

Therefore his new sale price will be $27\,\$$ because:

$$25+\frac{8}{100}\left(25\right )=27\,\$$$

So from this I thought that this can be used in the previous two relations.
But how?.

I tried to establish this inequation:

$26\left(8000+x\right)<25\left(8000\right)+27\left (8000+x\right)$

But that's where I'm stuck at since it is not possible to obtain a reasonable result from this as one will will be negative and the other positive.

The logic I used was to add up $8000\,\$$ plus something which is the production cost must be less than what has been obtained from selling the first $8000$ units plus a quantity to be added to those $8000$.

However there seems to be an error in this approach. Can somebody help me to find the right way to solve this problem?
Chemist116 is offline  
April 13th, 2018, 08:18 PM   #2
Senior Member
romsek's Avatar
Joined: Sep 2015
From: USA

Posts: 1,847
Thanks: 953

you are way overthinking it.

It's pretty clear he needs to sell more than 8000 tvs.

So the money he makes on them is

$i(k) = 25k + 8(k-8000)$

The cost to make them is

$c(k) = 26k$

Break even occurs when income equals cost so

$25k+8(k-8000) = 26k$

$8k - 640000 = k$

$k = \dfrac{64000}{7} = 9142.86 \rightarrow 9143$ as we can't sell a fraction of a set.

I don't see this correct answer listed.
romsek is online now  

  My Math Forum > High School Math Forum > Elementary Math

find, involving, number, objects, problem, profits

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
How do I find the number of elements in a problem involving least common multiples? Chemist116 Pre-Calculus 16 February 24th, 2018 01:05 PM
Using moments of inertia to find the heights of objects tankertert Calculus 1 January 2nd, 2014 01:47 AM
proof involving real number! eChung00 Real Analysis 2 March 12th, 2013 02:41 PM
I need help with maximizing profits problem BlahBlah1 Algebra 2 August 30th, 2010 09:52 AM
Maximizing Profits Problem matt1234 Calculus 7 January 3rd, 2010 03:41 PM

Copyright © 2018 My Math Forum. All rights reserved.