• 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 Application of Derivatives to EconomicsEconomics

Done By:

Mayas,Muneera,khloodMayas,Muneera,khlood

and Zahraand Zahra

1-Maximizing profit

Suppose that the demand equation for a monopolist’s product is p =400-2q the average –cost function is =0.2+4)+400/q),where q is number of units ,and both p and 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.

cc

• 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

p (x) is called marginal profit and it is a term

used in economics

Now p ̀=-4.4Is 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

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)

2-Minimizing the Cost

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 Here q must be positivepositive, to minimize C¯, we , to minimize C¯, we differentiate:-differentiate:-

(dc¯/dq)=¼+0+ ((q (0)-400(1))/q²)(dc¯/dq) =¼- (400/q²)To get the critical valuesLet (dc¯/dq) =0 (1/4)- (400/q²) =0 q²=1600 q=±√1600=±40 q= 40 Or q=-40

this is the rejectcritical number

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>0Which 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=0Divided by 3 will be:

-x²+300x+17500=0

(-x+350)(x+50)=0

-x=-350 or x=-50

X=350

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 divided by -2

q=40

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