Ahhh Applied Calc problem, i need an answer by 2am, thats when it is due HELP

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The applied calculus problem involves determining maximum profit from producing goods, where the cost function is given by 0.4q^2 + 10q and the selling price per item is $19. The profit function is established as profit = 19q - (0.4q^2 + 10q), simplifying to profit = 9q - 0.4q^2. To find maximum profit, the derivative of the profit function is taken and set to zero, which leads to the condition where marginal cost equals the selling price. The marginal cost is calculated as 0.8q + 10, and solving the equation 0.8q + 10 = 19 yields the optimal quantity to produce. This process ultimately allows for the calculation of income, costs, and profit.
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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)
 
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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)
 
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.
 
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.
 
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