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April 13th, 2018, 07:31 PM  #1 
Member Joined: Jun 2017 From: Lima, Peru Posts: 67 Thanks: 1 Math Focus: Calculus  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:
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 $$208000\$200000\$\,=\,8000\$$$ 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? 
April 13th, 2018, 09:18 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,324 Thanks: 1233 
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 $i(k) = 25k + 8(k8000)$ The cost to make them is $c(k) = 26k$ Break even occurs when income equals cost so $25k+8(k8000) = 26k$ $8k  640000 = k$ $k = \dfrac{64000}{7} = 9142.86 \rightarrow 9143$ as we can't sell a fraction of a set. I don't see this correct answer listed. 
July 9th, 2018, 08:06 PM  #3  
Member Joined: Jun 2017 From: Lima, Peru Posts: 67 Thanks: 1 Math Focus: Calculus  Quote:
But out of curiosity is this problem related with linear programming or something?.  

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