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olga_2 [115]
2 years ago
15

When you sell a slice for $4 you can sell a total of 300 slices. for every $0.10 increase in the price of a slice, the number of

slices you sell will decrease by 5. what price must you sell the slices for in order to maximize your revenue
Please send help ty
Mathematics
2 answers:
anygoal [31]2 years ago
8 0
At a price of $8 per unit, suppliers are willing to provide 6 units. How many units are consumers interested in purchasing at a price of $4 a unit? At eight units consumers are interested in purchasing at a price of $4 a unit.
Amiraneli [1.4K]2 years ago
8 0

Step-by-step explanation:

the original revenue is

4×300 = $1200

revenue(n) = (4 + n×0.1)×(300 - n×5) = 1200 - 20n + 30n - 0.5×n² =

-0.5×n² + 10n + 1200

with n being the number of units of $0.10 increasing the price.

now, where is the extreme (maximum, we hope) value of the function ?

it is the 0 point of the first derivative :

revenue'(n) = -n + 10 = 0

n = 10

so, the maximum is when increasing the price by 10 times of $0.10 (that means by increasing it by a full $1).

that means the best is selling the slices for 4+1 = $5.

and the max. revenue is then

(4+1)(300-50) = 5×250 = $1250.

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