How to find the least number of objects in a problem involving profits?

175 Views Asked by At

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 side 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?

2

There are 2 best solutions below

3
On

We know that the owner loses money for any production below $8000$ units, since it costs him more to produce his TV's than what he is receiving, so we consider production above $8000$ units.

Let's define $f:[8000,\infty)\rightarrow \mathbb{R}$ as follows: \begin{align} f(x)&= 25\cdot8000 + 27\cdot (x-8000)-26\cdot x\\ &=200000+27x-216000-26x\\ &=x-16000 \end{align} where $x$ represents the units sold. Notice that $25\cdot8000$ is the money he gets for the first $8000$ units sold, $27\cdot (x-8000)$ is the total money he gets for units sold after number $8000$, and $26x$ is the total amount of money it costs him to produce the units. So then $$f(x)=x-16000 > 0 \iff x > 16000.$$ To which we conclude that the owner has to sell at least $16,001$ units in order to make a profit.

0
On

He gets $25$ for the first $8000$ and $25\cdot 1.08=27$ for every one after that. Let him sell $n$. He receives $25\cdot 8000+27\cdot (n-8000)$ which has to be greater than $26 \cdot n$. These are equal at $n=16000$, so he needs to sell $16001$ to make a profit.