# Optimization price per unit

I have no idea how to do this, I tried a lot of things but they don't make sense and I have too many variables.

A manufacturer has been selling lamps at the price of \$6/lamp, and at this price they have been selling 3000 lamps a month. The manufacturer wishes to raise the price and estimated that for each \$1 increase they will sell 1000 fewer lamps a month. The manufacturer can produce the lamps at a cost of \$4 per lamp Express the manufacturers monthly profit as a function of the price that the lamps are sold, draw the graph and estimate the optimal selling point. I think the profit should be$\#(\mathrm{lamps\ sold})\cdot(\mathrm{price\ of\ lamps}) - 4\cdot\#(\mathrm{lamps\ sold})$. - ## 1 Answer If the manufacturer decides to set her price to \$$p then her estimation of the number of sold lamps per month is$$ 3000-(p-6)1000=9000-1000p. $$Hence the monthly profit is$$ (9000-1000p)(p-4)=-1000p^2+13000p-36000.$$Computing the derivative and setting it to zero shows that the optimal selling point is$p^*=6.5$. - I don't follow at all what happened, what that first formula is, where 900 came from or what anything else is after that. – user138246 Oct 22 '11 at 23:06 The$9000$comes from$3000-(-6)1000$. In the first formula I just started with$3000$and substracted$1000$for every dollar the price$p$is greater than$6\$. The second formula is the profit exactly as you suggest in the last line of your question. – Rasmus Oct 22 '11 at 23:10
I still don't understand how you got those numbers. – user138246 Oct 22 '11 at 23:13
@Jordan: I need to go to bed right now, but if you try to specify what's unclear to you then I can add some explanations later. – Rasmus Oct 22 '11 at 23:20
I give up, I just can't do word problems and I am wasting my time trying to study these. There are other parts of the test I will fail if I don't study that, thanks though. – user138246 Oct 22 '11 at 23:22