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• The major four points that every one in economic interested about are: Maximizing the Profit (Mayas) Minimizing the Average Cost (Muneera) Maximizing the Revenue (Zahra) Price Elasticity of Demand (Klood) Application of Derivatives to Economics Done By: Mayas,Muneera,khlood and Zahra 1-Maximizing profit Suppose that the demand equation for a monopolist’s product is p =400-2q the average – cost function is c =0.2+4+)400/q),where q is number of units ,and both p and c are expressed in dollars per units. a-Determine the level of output at which profit is maximized. b-Determine the price at which maximum profit occurs. c- Determine the maximum profit. • Solution: we know that profit =t revenue –t cost Since total revenue r and total cost c are given by r =pq = 400q -2q2 And c=qc = 0.2q2-4q+400 , The profit is: p=r- c= 400q -2q2_ (0.2q2-4q+400) =360q – 2.2q2 -400……………..)1) a-To maximize profit, let p ̀ =0 p ̀=369 -4.4q=0 q =90 p̀(x) is called marginal profit and it is a term used in economics Now p ̀=-4.4 Is always negative , so it is negative at the critical value q=90. By the second-derivative test then, there is a relative maximum there .Since q= 90 is the only critical value on (0,),we must have an absolute maximum there. b- the price at which maximum profit occurs is obtained by setting q =90 in the demand equation: p= 400 -2(90) =$220. C- the maximum profit is obtained by replacing q by 90 in Eq (1) which gives p= $14,180 2-Minimizing the Cost To study the effect of production levels on cost ,economics use the average cost function C¯ defined as: C¯ =C/x where C(x) is the cost of producing x units of a certain product The marginal cost function is the derivative ,C`(x) Example: A manufactures total cost function is C= (q²/4) +3q+400 were C is the total cost of producing q units. At what level of output will average cost per unit be a minimum? What is this minimum? Solution: The quantity to be minimized is the average cost C¯= (c/q) = (¼q²+3q+400)/q Where C is the cost and q is the product C¯=¼(q²/q) + (3q/q) + (400/q) =¼q+3+ (400/q) (1) Here q must be positive, to minimize C¯, we differentiate:- (dc¯/dq)=¼+0+ ((q (0)-400(1))/q²)(dc¯/dq) =¼- (400/q²) To get the critical values Let (dc¯/dq) =0 (1/4)- (400/q²) =0 q²=1600 q=±√1600=±40 q= 40 Or q=-40 this is the critical number reject To determine whether this level of output gives a relative minimum, we shall use the second derivative test:- (d²c¯/dq)= q² (0)-400+2q/(q²)² =800/q³ >0 (d²c¯/dq)(40)= 800/(40)³=0.0125>0 Which is positive for q=40. Thus, C¯ has a relative minimum when q=40. Since q=40 is the only relative ex-tremum, we conclude that this relative minimum is indeed an absolute minimum. There is a minimum value to average cost at q= 40 Substituting q=40 in equation (1) C¯ (40) =¼(40) +3+ (400/40) = 10+3+10=23 Example (2):A company estimates that cost of producing x units of a certain product is given by C= 800+0.04x+0.0002x² find the production level that minimizes the average cost per unit. Compare this minimal average cost to the average cost when 400 units are produced ? 1-We let x be the number of units produced 2-The primary equation represent the quantity to be minimized is : C¯=C/x 3-Substituting from the given equation for C, we will have: C¯= (800+0.04x+0.0002x²/x) = (800/x)+0.04+0.0002x 4-Setting the derivative =0 (dc¯/dx)=-(800/x²) +0.0002=0 x²= (800/0.0002)= 4000000 x= 2000 For 2000 units the average cost is: C¯= (800/2000) _0.04+0.0002(2000) =0.84$ For 400 units the average cost is: C¯= (800/400) +0.04+0.0002(400) =2.12 $ 3-maximizing the revenue Revenue in economics means: the amount of money that a company earns from its activities in a given period, mostly from sales of products and/or services to customers. Example:Suppose a company has determined that its total revenue R for a certain product is given by: R= -x³+450x²+52500x where R is measured in dollars and x is the number of units produced. What production level will yield maximum revenue? Solution: 1-Sketch is given. 2- R= -x³+450x²+52500x 3-Since R is already given as a function of one variable, we do not need a secondary equation. 4-Differentiating and setting the derivative =0 (dR/dx)= -3x²+900x+52500=0 Divided by 3 will be: -x²+300x+17500=0 (-x+350)(x+50)=0 -x=-350 or X=350 x=-50 The critical value are x =350 and x = -50. Choosing the positive value of x, we conclude that the maximum revenue is obtained when 350 units are produced. Example (2):- The demand equation for a manufactures product is: P= (80-q)/4, 0≤q≤80 Where q is the number of units and P is the price per unit. At what value of q will there be maximum revenue? What is the maximum revenue? Let R be the revenue, which is the quantity to be maximized. Since Revenue= (price). (Quantity) We have r = pq= ((80-q)/4).q= ((80q-q²)/4) Where 0≤q≤80 Setting (dR/dq) =0,we obtain (dR/dq)= (80-2q)/4=0, multiply by4 (dR/dq)= 80-2q=0 -2q=-80 q=40 divided by -2 Thus, 40 is the only critical value. Now we see whether this gives a maximum. Examining the second derivative : (d²r/dx)= -1/2 we conclude that q=40 gives the absolute maximum revenue, namely, [80(40)-(40) ²]/4=400 Is an economists way that describe the behavior of a demand function. It describes the relative responsiveness of consumers to a change in the price of an item. if x=f(p) is a differentiable demand function, then the price elasticity of demand is given by η= (p/x)(dx/dp) (where η is the greek lowercase letter eta).For a given price, if l η l<1, the demand is said to be inelastic; if l η l>1, the demand is said to be elastic. Show that the demand function , x= 2500/P^2, is elastic. Solution η= ( p/x ) ( dx / dp ) = ( p/(2500/p^2)) [ (-2)(2500)/p^3] = (-5000 p^3) / 2500 p^3 = -2 since n = 2, we conclude that the demand is elastic for any price. Show that for a differentiable demand function, the marginal revenue is positive when the demand is elastic and negative when the demand is inelastic. (assume that the quantity demanded increases as the price decrease and thus dx/dp negative Solution: since the revenue is guven by R=xp, we calculate the marginal revenue to be (dR/dx)= x(dp/dx) + p = p [ (x/p) (dp/dx) + 1 ] = p[1/( (x/p) (dp/dx)) +1] = p( (1/n)+1) If the demand is inelastic, then lnl<1 implies that 1/n <-1,since dx/dp is negative.( We assume that x and p are positive.) Therefore, (dR/dx) = p((1/n) + 1) Is negative. Similarly,if the demand is elastic, then lnl >1 and -1<1/n, which implies that dR/dx is positive