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 April 13th, 2018, 06:31 PM #1 Member     Joined: Jun 2017 From: Lima, Peru Posts: 40 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, 08:18 PM #2 Senior Member 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) = 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.

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