# Derivatives, Second Derivatives and Profit Function

A. Write a function for your profits for each price you charge. This is done by multiplying (P-.5) times your function (y= -100x + 250). I.e. if your function is Cups Sold = 1000 - 100P, your profit function would be (P - .5)*(1000 - 100P).

B. Calculate the first derivative of your profit function, and create another table with the price, profit, and value of the first derivative at the prices below. Can you tell what your profit maximizing price is from this table?

Price x Profit the derivative

0.25 -56.25 250

0.5 0 200

0.75 43.75 150

1 75 100

1.25 93.75 50

1.5 100 0

1.75 93.75 -50

2 75 -100

2.25 43.75 -150

2.5 0 -200

C. Calculate the second derivative, and also use the first derivative to find the profit maximizing price. What is the price, and what does the second derivative tell you?

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#### Solution Summary

Derivatives, Second Derivatives and Profit Function are investigated for a lemonade stand. The solution is detailed and well presented.

Profit Function and Maximum Profit

A manufacturer finds that the total profit from producing and selling Q units of a product is given by the profit function:

Total Profit = f(Q) = - 460 + 100Q - Q^2

1. Compute the value of the function at Q=10

Total Profit = f(10)= - 460 + 100(10) - 10^2

Total Profit = f(10)= - 460 + 1000 - 100

Total Profit = f(10)=540 - 100

Total Profit = f(10)=440

2. Compute the value of the first derivative of the function at Q=10

3. Explain the significance of each computation.

4. At what level of Q is Profit equal to 1,815?

5. Use Calculus: At what level of Q will Total Profit be a maximum?

6. Double-check your answer to part 5 with an Excel graph.

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