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An ice cream shop finds that its weekly profit P (measured in dollars) as a function of the price x (measured in dollars) it charges per ice cream cone is given by the function k, defined by k(x) = 125x^2 +670x 125 where P= k(x).

Required:
a. Determine the maximum weekly profit and the price of an ice cream cone that produces that maximum profit
b. The cost of the ice cream cone is too low then the ice cream shop will not make a profit. Determine what the ice cream shop needs to charge in order to break even
c. The profit function for Cold & Creamy (another ice cream shop) is defined by the function g where gx)- k(x-2). Does the function g have at the same maximum value as k?

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Answer:

The answer is below

Step-by-step explanation:

Given that k(x) = -125x² +670x - 125, where x is the charge per cone and P = k(x) is the weekly profit

a) The maximum profit is at P'(x) = 0. Therefore we have to find the derivative of the profit equation and equate it to 0.

P'(x) = k'(x) = 0

k'(x) = -250x + 670 = 0

-250x + 670 = 0

250x = 670

x = $2.68

P = k(2.68) = -125(2.68²) + 670(2.68) - 125 = $772.8

Hence the maximum profit is $772.8 when the price of each ice cream cone is $2.68

b) At break even, the profit is 0. Hence P= k(x) = 0

-125x² + 670x - 125 = 0

x = 5.17 or x = 0.19

Therefore to break even, the price of the ice cream cone needs to be $0.19 or $5.17

c) g(x) = k(x - 2)

g(x) = -125(x - 2)² + 670 (x -2) - 125

Maximum profit is at g'(x) = 0

g'(x) = -250(x-2) + 670

-250(x-2) + 670 = 0

-250x + 500 + 670 = 0

250x = 1170

x = 4.68

g(4.68) = -125(4.68 - 2)² + 670(4.68 -2) - 125 = $772.8

Therefore function g has the same maximum value as function k