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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: 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$$208000\$-200000\$\,=\,8000\$$So I thought what If I consider the second part of the problem which it says that he will receive an additional of 8% after 8000 units. Therefore his new sale price will be 27\,\$$ because: $$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$i(k) = 25k + 8(k-8000)$The cost to make them is$c(k) = 26k$Break even occurs when income equals cost so$25k+8(k-8000) = 26k8k - 640000 = kk = \dfrac{64000}{7} = 9142.86 \rightarrow 9143$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:  Originally Posted by romsek 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) = 26k8k - 640000 = kk = \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.
Sorry about the late reply romsek. Indeed your approach was what I should have done. My apologies but there was an error in the problem when I transcribed it. It should have said 8% more not just 8. By doing this change in your calculation should had been 27 and with that adjustment then the answer appears in the alternatives given. Which should had been 16001.

But out of curiosity is this problem related with linear programming or something?.

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