Explanation Given information: In order to sell x refrigerators, the price per refrigerator must be: p = 280 − 0.4 x . The provided cost function is: C ( x ) = 5000 + 0.6 x 2 Formula used: The concept of Max-Min Principle-2 is taken into consideration which is as follows: Suppose f is a function such that f ' ( x ) exists for every x in an interval l and there is exactly one critical value c in l , for which f ' ( c ) = 0 Then, f ( c ) is absolute maximum value over l if f ' ' ( c ) < 0 Or f ( c ) is absolute minimum value over l if f ' ' ( c ) > 0 Proof: As it is provided that in order to sell x refrigerators, the price per refrigerator must be: p = 280 − 0.4 x . Now, the revenue generated for selling x items will be obtained by: R ( x ) = p ⋅ x Substitute 280 − 0.4 x for p in equation R ( x ) = p ⋅ x to get: R ( x ) = x ( 280 − 0.4 x ) Thus, the total revenue is R ( x ) = x ( 280 − 0.4 x ) . Consider the revenue function is: R ( x ) = x ( 280 − 0.4 x ) And, the provided cost function is: C ( x ) = 5000 + 0.6 x 2 Now, the profit function is: P ( x ) = R ( x ) − C ( x ) Thus, the profit function will be: P ( x ) = R ( x ) − C ( x ) = 280 x − 0.4 x 2 − ( 5000 + 0.6 x 2 ) = − x 2 + 280 x − 5000 Thus, the required profit function is P ( x ) = − x 2 + 280 x − 5000 . Consider the profit function is: P ( x ) = − x 2 + 280 x − 5000 Now, first find the derivative of above function with respect to x : P ′ ( x ) = − 2 x + 280 Now, as x is the number of refrigerators which cannot be negative, thus the derivative exists for all real numbers in the interval [ 0 , ∞ ) . Now, solve P ′ ( x ) = 0 as: P ′ ( x ) = 0 − 2 x + 280 = 0 − 2 x = − 280 x = 140 Since, there is only one critical value, use the second derivative to determine whether it is a maximum as: P ″ ( x ) = − 2 < 0 The second derivative is negative for all real values of x

Calculus and Its Applications (11th Edition)

11th Edition

ISBN: 9780321979391

Author: Marvin L. Bittinger, David J. Ellenbogen, Scott J. Surgent

Publisher: PEARSON

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Chapter 2.5, Problem 84E

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To prove: That the maximum profit for the profit function P(x)=−x2+280x−5000 is $14,600.

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Chapter 2.5, Problem 83E

Chapter 2.5, Problem 85E

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To prove: That the maximum profit for the profit function P ( x ) = − x 2 + 280 x − 5000 is $ 14 , 600 .

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To prove: That the maximum profit for the profit function P(x)=−x2+280x−5000 is $14,600.

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