You want to sell a certain number n of items in order to maximize your proﬁt. Market research tells you that if you set the price at $1.50, you will be able to sell 5000 items, and for every 10 cents you lower the price below $1.50 you will be able to sell another 1000 items. Suppose that your ﬁxed costs (“start-up costs”) total $2000, and the per item cost of production (“marginal cost”) is $0.50.

Solve the above engineering problem using calculus.

Hints: Find the price to set per item and the number of items sold in order to maximize proﬁt, and also determine the maximum proﬁt you can get.

To achieve M3, you must present your work using coherent, logical development of principles/concepts in determining the maximum profit and use technical language accurately.
To achieve D1, you must evaluate the validity of your answer using the given data.

Let P(x) be set for the profit since the price per item is x. Profit minus cost, profit is the number of items sold multiplied by the price per. P=nx-2000-0.50n, n=5000+1000(1.5-x)/0.10
P(x)=(5000+1000(1.5-x)/0.10)x-2000-0.5(5000+1000(1.5-x/0.10)=10000x²+25000x-12000.
If x=1.25 what is the gross value, and the critical value. P(0)=12000, P(1.25=3625 and P(1.25)=3000, note P(1.25 is the price set). By selling the items for $1.25 we have a profit of $3625 when 7500 items are sold.

francocanacari francocanacari · Answer 2 · 2021-11-04T00:54:48+01:00

The price that would maximize profit is $ 1.25, with which 7,500 items will be sold and a profit of $ 3,635 will be collected.

Since you want to sell a certain number of items in order to maximize your profit, and if you set the price at $ 1.50, you will be able to sell 5000 items, and for every 10 cents you lower the price below $ 1.50 you will be able to sell another 1000 items, supposing that your ﬁ xed costs total $ 2000, and the per item cost of production is $ 0.50, to find the price to set per item and the number of items sold in order to maximize pro ﬁ t, and also determine the maximum proﬁt you can get, the following calculations must be performed:

(5000 x 1.50) - (5000 x 0.50) - 2000 = 3000
(6000 x 1.40) - (6000 x 0.50) - 2000 = 3400
(7000 x 1.30) - (7000 x 0.50) - 2000 = 3600
(8000 x 1.20) - (8000 x 0.50) - 2000 = 3600
(7500 x 1.25) - (7500 x 0.50) - 2000 = 3625

Therefore, the price that would maximize profit is $ 1.25, with which 7,500 items will be sold and a profit of $ 3,635 will be collected.

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