Applications of the Derivative Chapter 14

Ch. 14 Applications of the Derivative 14.1 Absolute Extrema 14.3 Business Applications of Extrema

14.1 Absolute Extrema ABSOLUTE MAXIMUM OR MINIMUM Let f be a function defined on some interval. Let c be a number in the interval. Then f(c) is the absolute maximum of f on the interval if f(x)  f(c) f(x)  f(c) A function has an absolute extremum (plural: extrema) at c if it has either an absolute maximum or absolute minimum there. for every x in the interval, and f(c) is the absolute minimum of f on the interval if for every x in the interval.

Absolute Extrema Consider the graph of function f on the interval [-4, 4] Absolute maximum at x = 4 (end point) Absolute minimum at x = 1

Absolute Extrema EXTREME VALUE THEOREM A function f that is continuous on a closed interval [ a, b ] will have both an absolute maximum and an absolute minimum on the interval. Each of these occurs either at an endpoint of the interval or at a critical number of f.

Absolute Extrema To find absolute extrema for a function f continuous on a closed interval [ a, b ] 1.Find all critical numbers for f in ( a, b ) 2.Evaluate f for all critical numbers in ( a, b ) 3.Evaluate f for the endpoints a and b of the interval [ a, b ] 4.The largest value found in step 2 or 3 is the absolute maximum for f on [ a, b ], and the smallest value found is the absolute minimum for f on [ a, b ] FINDING ABSOLUTE EXTREMA

Absolute Extrema – example (closed interval) Find the absolute extrema of the function f(x) = 2 x x 2 – 12 x – 7 on the interval [-3, 0] Solution: 1.Find critical numbers f ‘(x) = 6 x x – 12 = 6( x + 2)( x – 1) Set f ‘(x) = 0 and solve for x Critical numbers are x = -2 and x = 1 Only -2 lies in the interval [-3, 0] 2, 3. Evaluate f for x = -2, and for endpoints x = -3 and x = 0. f (-2) = 13 f (-3) = 2 f (0) = -7 4.Absolute maximum occurs at (-2, 13), Absolute minimum occurs at (0, -7)

Absolute Extrema – example (closed interval) (-3, 2) Absolute maximum (-2, 13) (0, -7) f(x) = 2 x x 2 – 12 x – 7 [-3, 0] Absolute minimum

Application For several weeks, the highway department has been recording the speed of freeway traffic flowing past a certain downtown exit. The data suggest that between 1:00 and 6:00 P.M. on a normal weekday, the speed of the traffic at the exit is approximately S(t) = t 3 – 10.5 t t + 20 mph where t is the number of hours past noon. At what time between 1:00 and 6:00 P.M. is the traffic moving the fastest, and at what time is it moving the slowest? In other words, find the absolute maximum and absolute minimum of the function S(t) on the interval [1, 6].

Application, solution 1. Find the critical numbers S’(t) = 3 t 2 – 21 t + 30 = 3( t 2 – 7 t + 10) = 3( t – 2)( t – 5) Set S’(t) = 0 t = 2, t = 5 2. Evaluate S for t = 2 and t = 5, and for endpoints t = 1 and t = 6. S (1) = 40.5 S (2) = 46 S (5) = 32.5 S (6) = 38 Absolute maximum = 46. Traffic is moving fastest at 2:00 Absolute minimum = Traffic is moving slowest at 5:00 S(t) = t 3 – 10.5 t t + 20 mph, [1, 6]

Application, solution S(t) = t 3 – 10.5 t t + 20 mph [1, 6] Maximum speed (2, 46) Minimum speed (5, 32.5)

14.3 Business Applications of Extrema 1.Economic lot size 2.Economic order quantity 3.Elasticity of Demand 4.Profit Maximization

1.Economic Lot Size Economic lot size – The quantity of output that allows a manufacturer to meet the demand for their product while minimizing total cost.  Total costs: Set-up costs Manufacturing costs Storage costs

Economic Lot Size Gadget, Inc. produces 12,000 widgets per year (annual demand = 12,000). The widgets can be produced in several batches of equal size per year.  Small number of large batches Reduces set-up costs. Increases storage costs.  Large number of small batches Increases set-up costs Reduces storage costs Economic lot size – The number of widgets that should be manufactured in each batch to minimize total cost.

Lowest set-up costs Highest storage costs Highest set-up costs Lowest storage costs

Economic Lot Size Let:  q = number of units in each batch;  k = cost of storing one unit of the product for one year;  f = fixed setup cost to manufacturer the product;  g= cost of manufacturing a single unit of the product;  M = total number of units produced annually Manufacturing cost per batch = f + gq The number of batches per year = M/q Total annual manufacturing cost =

Economic Lot Size Cost of storing one unit for a year = k. Average inventory = q /2 Total storage cost = Let:  q = number of units in each batch;  k = cost of storing one unit of the product for one year;  f = fixed setup cost to manufacturer the product;  g= cost of manufacturing a single unit of the product;  M = total number of units produced annually

Economic Lot Size Total production cost = manufacturing cost + storage cost To find the value of q that minimizes T(q), find T’(q) and set T’(q) = 0 (Remember: f, g, k, and m are constants) Set T’(q) = 0 solve for q

Economic Lot Size To find the value of q that minimizes T(q), find T’(q) and set T’(q) = 0 (Remember: f, g, k, and m are constants)

Economic Lot Size INVENTORY PROBLEM Economic lot size that minimizes total production cost Where q = quantity, f = fixed setup cost, M = total units produced annually, k = cost of storing one unit for one year.

Economic Lot Size - example Suppose 100,000 lamps are to be manufactured annually. It costs $1 to store a lamp for 1 year, and it costs $500 to set up the factory to produce a batch of lamps. 1. Find the number of lamps to produce in each batch. 2. Find the number of batches of lamps that should be manufactured annually.

Solution 1. k = 1, M = 100,000, f = 500

2.Economic Order Quantity Deals with the problem of reordering an item that is used at a constant rate throughout the year.  How often to order?  How many units to order each time an order is placed? q = number of units to order each time k = cost of storing one unit for one year f = fixed cost to place an order M = total units needed per year Goal: Minimize the total cost of ordering over a year’s time, where Total cost = Storage cost + Reorder cost

Economic Order Quantity As defined previously, average inventory = q /2, so yearly storage cost is kq /2 Number of orders placed annually = m/q, so reorder cost is fM/q Total reorder cost is and, When T’(q) is set = 0 and, we solve for q Same as the “inventory problem”

Example A bookstore has an annual demand for 100,000 copies of a best-selling book. It costs $.50 to store 1 copy for 1 year, and it costs $60 to place an order. Find the optimum number of copies per order. Solution k = 0.5, M = 100,000, f = 60

3.Elasticity of Demand The responsiveness of the quantity demanded of a good to a change in its price, ceterus paribus. ELASTICITY OF DEMAND Let q = f(p), where q is quantity demanded at a price p. The elasticity of demand is Demand is inelastic if E < 1 Demand is elastic if E > 1 Demand has unit elasticity if E = 1

Example Research has indicated that the demand for heroin is given by q = 100 p Find E 2. Is the demand for heroin elastic or inelastic? 0.17 < 1, the demand is inelastic.

Elasticity of Demand For many products, elasticity varies at different price levels. Revenue and Elasticity If demand is inelastic, total revenue increases as price increases. If demand is elastic, total revenue decreases as price increases. Total revenue is maximized at the price where demand has unit elasticity.

Example Assume the demand for a product is, where p is the price in dollars. a. Find the price intervals where demand is elastic and where demand is inelastic.

Example Assume the demand for a product is, where p is the price in dollars. a. Find the price intervals where demand is elastic and where demand is inelastic.  Solution Since, and

Example Assume the demand for a product is, where p is the price in dollars. a. Find the price intervals where demand is elastic and where demand is inelastic. To decide where E 1, solve the equation E = 1 At p = $6, demand is unit elastic.

Example Assume the demand for a product is, where p is the price in dollars. a. Find the price intervals where demand is elastic and where demand is inelastic. Substitute a test number on either side of 6 in the expression for E to see which values make E 1 The demand is inelastic at prices less than $6, and elastic at prices greater than $6

Profit Maximization Marginal Analysis Criterion For Maximum Profit Profit P(q) = R(q) – C(q) is maximized at a level of production q where marginal revenue equals marginal cost; that is, where R’(q) = C’(q)

D MC ATC MR Profit MR = MC Profit Per Unit Q Price, costs, and revenue Remember the MR=MC Rule?

Profit Maximization - Example A manufacturer determines that when q thousand widgets are produced (and sold each month), the total revenue will be R(q) = q q and the total cost will be C(q) = 0.4 q q + 40 What level of production will maximize profit? Or, at what level of production does MR = MC ? MR = R’(q) = -2.4 q MC = C’(q) = 0.8 q q = 0.8 q q = 19.2 q = 6

Profit Maximization - Example Perform the first derivative test:  P(q) = MR – MC = (-1.2 q q ) – (0.4 q q – 40) = -1.6 q q – 40 P’(q) = q P’ (6) = - 3.2(6) = 0 P’ (5) = 3.2 (increase q to increase P ) P’ (7) = -3.2 (decrease q to increase P )

P(q) = -1.6 q q – 40 Max profit at q = 6

Chapter 14 End 