Maximum profit with cost function 0.4q² + 10q and selling price $19

[JasonJo]

Ahhh! Applied Calc problem, i need an answer by 2am, that's when it is due! HELP!

the cost of producing q goods = .4q^2 + 10q

the marginal cost is given by the derivative, .8q + 10

what is the maximum profit if each item is sold for 19 dollars (assume you sell everything you produce)

 

[Duarh]

set up the profit function (income minus costs) and then take the derivative of the whole thing and set it equal to zero. looking at marginal cost separately is not that useful (though, of course, by taking the derivative of the whole thing and setting it equal to zero you'll be looking for that magic point where the marginal cost and marginal profit are equal)

 

[wisky40]

profit=price of producing q goods-cost of producing q goods =>

profit= 19(q)-(.4q^2 +10q) =9(q)-.4q^2 from here you need either to find the vertex of this parabola or to derivate for its maximum.

 

[HallsofIvy]

Since you are given the marginal cost, the maximum profit occurs when the price of the item equals the marginal cost. (Producing one more would make the cost greater than the price and you start losing money.)

That is .8q+ 10= 19. Solve that for q and then find:
1) income: 19q
2) cost .4q^2 + 10q
3) Subtract to find profit.