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# Week 7 of Lecture Notes MATH127

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This 4 page Class Notes was uploaded by Monica Weisenbach on Saturday March 28, 2015. The Class Notes belongs to MATH127 at University of Massachusetts taught by Thurlow Cook in Spring2015. Since its upload, it has received 38 views. For similar materials see Calculus I for Life and Social Science Majors in Mathematics (M) at University of Massachusetts.

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Date Created: 03/28/15

Week 7 Lecture Notes Minimums and Maximums Remember minimums are low points maximums are high points Local minsmaxes relative minsmaxes Global minsmaxes absolute minsmaxes The roots of the first derivative give the critical points of the graph where the graph has minimums or maximums A sign change of positive to zero to negative is a maximum A sign change of negative to zero to positive is a minimum The roots of the second derivative give the inflection points of the graph where the graph changes concavity Example 1 fx 3x44x36 f x 12x312x2 gt 12x2x1 roots are 01 f x 36x224x gt 12x3x2 roots are 0 23 x 0 23 1 fX 6 54 5 f x o 17 o f X O 0 12 Charts like this are helpful by tracking the signs of the first and second derivatives Do this by testing numbers between the interval From this we can see that The concavity changes twice at O and 23 There is a minimum at 1 because the sign changes from to O to The critical point at 0 turns out to be nothing Try using this to draw the graph Example 2 fx X36X24X2 f x x22x21 roots are 21 f x x12 roots are 12 x 2 12 1 fx 36 25 5 f x O O625 O f x 15 O 12 From this we see that There is a maximum at 2 and a minimum at 1 There is an inflection point at 12 Try to draw the graph Example 3 fx x33x29x15 domain 54 watch the domain f x 3X26X9 gt 3X2 2X3 gt 3X3X1 roots are 1 3 f x 6x6 gt 6x1 roots are 1 x 5 1 1 3 fx 140 20 4 12 f x O o W o With the domain don t forget to Check the endpoints With this we see There is a maximum at 1 and a minimum at 3 There is an inflection point at 1 Try drawing the graph Graphs for the above check your work Example 1 Example 2 Example 3 Revenue Functions Revenue Rq is expressed as a function of quantity Rq pq revenue prioequantity Cost of producing the quantities Cq is also expressed as a function of quantity This has a fixed cost investment cost Profit revenue cost of producing items aka P RqCq Marginal revenue is the derivative of the revenue function R Marginal cost is the derivative of the cost function 0 How does one maximise profit Profit is maximised when marginal P canQ l In l39 a 25 I F RG Cql Pra f d39 1 011 a If a1 gt R q loss revenue R marginal cost 0 aka when the rate of money coming in the rate of money going out Example An ice cream company can sell 4000 units at 4 With a 025 decrease the number of units sold increases by 200 units Find the price and quantity that maximize revenue Rq pq 44000 16000 APAq O252OO 1800 locally the graph can be treated as a linear line In y mxb form p 1800qb p 18004000b b9 So p 1800q9 gt put this in the revenue function Rq pq 1800q9q q28009q R q 2q8009 q4009 set to zero Set to zero and q3600 Go back to p 1800q9 and p 180036009 45 So the answer is 3600 units at 450 Logistic Curve Growth Functions The equation is L1Cekt LC and k gt O and are constants L is the horizontal asymptote the upper limit L2 is the inflection point of the curve k affects the steepness of the curve and is the growth rate L P L1Cekt gt P1CektL gt PkCe39 kt Ce39ktP 0 L P kCektP1Cekt amp r if 1quot quotquotquot

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