d i E Applied Maxima and Minima a l l u d Dr. Abdulla Eid b - PowerPoint PPT Presentation

Section 13.6 d i E Applied Maxima and Minima a l l u d Dr. Abdulla Eid b A College of Science . r D MATHS 104: Mathematics for Business II Dr. Abdulla Eid (University of Bahrain) Applications 1 / 17

Example A company manufacturer sells q phones per week. The weekly demand is given by p = 500 − 0.5 q Find the selling price that maximizes the revenue. d i E Solution: a We find the revenue function to be l l u d b r = qp A We find the derivatives first which are . r r ′ = p + qp ′ D r ′ = 500 − .5 q + q ( .5 ) r ′ = 500 − .5 q + q ( − .5 ) r ′ = 500 − q r ′′ = − 1 Dr. Abdulla Eid (University of Bahrain) Applications 2 / 17

Critical Points To find the critical points, we find where the first derivative equal to zero d or does not exist. i E a f ′ ( x ) does not exist l l u r ′ ( x ) = 0 denominator = 0 d b numerator = 0 1 = 0 A 500 − q = 0 Always False . r q = 500 D No Solution Dr. Abdulla Eid (University of Bahrain) Applications 3 / 17

Maximum or Minimum d i To find out whether q = 500 is a maximizer or minimizer, we will use the E second derivative test. so we check a l l u r ′′ ( 500 ) = − 1 < 0 d b A So q = 500 is a local maximizer with maximizer price p = 500 − 0.5 ( 500 ) = 250 and maximum revenue r ( 500 ) = 125000. . r D Dr. Abdulla Eid (University of Bahrain) Applications 4 / 17

Exercise (Old Final Exam Question) A company manufacturer sells q phones per week. The weekly demand is given by p = − 50 q + 300 d i E Find the selling price that maximizes the revenue. a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Applications 5 / 17

Example (Old Final Exam Question) For a certain product, the demand function is p = 100 e − 0.01 q , where q is the number of items. Find the selling price that maximizes the revenue. d Solution: i E We find the revenue function to be a l r = qp = q ( 100 e − 0.01 q ) l u d b We find the derivatives first which are A r ′ = p + qp ′ . r r ′ = 100 e − 0.01 q + 100 qe − 0.01 q ( − 0.01 ) D r ′ = 100 e − 0.01 q − qe − 0.01 q r ′ = ( 100 − q ) e − 0.01 q = 100 − q e 0.01 q r ′′ = − e − 0.01 q + ( 100 − q ) e − 0.01 q ( − 0.01 ) Dr. Abdulla Eid (University of Bahrain) Applications 6 / 17

Critical Points To find the critical points, we find where the first derivative equal to zero d or does not exist. i E a f ′ ( x ) does not exist l r ′ ( x ) = 0 l u denominator = 0 d numerator = 0 b e 0.01 q = 0 A ( 100 − q ) = 0 0.01 q ∗ = ln 0 = − ∞ . q − 100 = 0 r D No Solution q = 100 Dr. Abdulla Eid (University of Bahrain) Applications 7 / 17

Maximum or Minimum d i To find out whether q = 100 is a maximizer or minimizer, we will use the E second derivative test. so we check a l l u d r ′′ ( 100 ) = < 0 b A So q = 100 is a local maximizer with maximizer price p = 100 e − 0.01 ( 100 ) = 100 e − 1 and maximum revenue r ( 100 ) = . . r D Dr. Abdulla Eid (University of Bahrain) Applications 8 / 17

Example (Old Exam Question) Assume the total cost of producing q units of a product is given by d i E c = q 3 − 24 q 2 + 250 q + 338 a l l Find the level of production that minimizes the cost. u d b Solution: A We find the derivatives first which are . r D c ′ = 3 q 2 − 48 q + 250 c ′′ = 6 q − 48 Dr. Abdulla Eid (University of Bahrain) Applications 9 / 17

Critical Points To find the critical points, we find where the first derivative equal to zero d or does not exist. i E a c ′ ( x ) does not exist l l u c ′ ( x ) = 0 denominator = 0 d b numerator = 0 1 = 0 A 3 q 2 − 48 q + 250 = 0 Always False . r D No Solution No Solution Dr. Abdulla Eid (University of Bahrain) Applications 10 / 17

Maximum and Minimum d i E a l Since there is no local minimum, we take q = 0 as a minimizer with l u minimum cost c = c ( 0 ) = 338. d b A . r D Dr. Abdulla Eid (University of Bahrain) Applications 11 / 17

Example (Old Exam Question) A manufacturer produces water bottles that sell for 400 BD each. The total cost of producing q bottles is given by the function c ( q ) = 0.012 q 2 − 29.6 q + 5080. What is the number of bottles d that should be made to maximize the profit. i E a Solution: l l We find the net profit function N ( q ) to be u d b N = Total Revenue − Total Cost = 400 q − ( 0.012 q 2 − 29.6 q + 5080 ) A . r D We find the derivatives first which are N ′ = 400 − 0.024 q + 29.6 N ′′ = − 0.024 Dr. Abdulla Eid (University of Bahrain) Applications 12 / 17

Critical Points To find the critical points, we find where the first derivative equal to zero d or does not exist. i E a N ′ ( x ) does not exist l l u N ′ ( x ) = 0 denominator = 0 d b numerator = 0 1 = 0 A 400 − 0.024 q + 29.6 = 0 Always False . r q = 17900 D No Solution Dr. Abdulla Eid (University of Bahrain) Applications 13 / 17

Maximum or Minimum d i E To find out whether q = 17900 is a maximizer or minimizer, we will use a the second derivative test. so we check l l u d N ′′ ( 17900 ) = < 0 b A So q = 17900 is a local maximizer with maximum net profit N ( 17900 ) = . . r D Dr. Abdulla Eid (University of Bahrain) Applications 14 / 17

Example (Old Exam Question) The demand function for a product is p = 700 − 2 q and the average cost per unit for producing q units is c ( q ) = q + 100 + 1000 q . Find the selling price that maximizes the profit and find the maximum profit. d i E Solution: a We find the net profit function N ( q ) to be l l u d N = Total Revenue − Total Cost = qp + qc ( q ) b A = 700 q − 2 q 2 − q 2 − 100 q − 1000 . = − 3 q 2 + 600 q − 1000 r D We find the derivatives first which are N ′ = − 6 q + 600 N ′′ = − 6 Dr. Abdulla Eid (University of Bahrain) Applications 15 / 17

Critical Points To find the critical points, we find where the first derivative equal to zero d or does not exist. i E a N ′ ( x ) does not exist l l u N ′ ( x ) = 0 denominator = 0 d b numerator = 0 1 = 0 A − 6 q + 600 = 0 Always False . r q = 100 D No Solution Dr. Abdulla Eid (University of Bahrain) Applications 16 / 17

Maximum or Minimum d i To find out whether q = 100 is a maximizer or minimizer, we will use the E second derivative test. so we check a l l u N ′′ ( 100 ) = − 6 < 0 d b A So q = 100 is a local maximizer with a maximizer price p = 500 and maximum net profit N ( 100 ) = 29000. . r D Dr. Abdulla Eid (University of Bahrain) Applications 17 / 17