Done By:
and
Mayas Mahmoud
Muneera AL-Saai
kholod Mustafa
Zahra
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Introduction:
There are a lot of applications of derivative in many scientific fields and our major
concern here in the area of economics.
The major four points that is every one in economic interested about are:
1) Maximizing the profit
2) Minimizing the cost
3) Price elasticity of Demand
The Good news, all of these four are described by a function, and we know by using
specific techniques of math ,we can come up of these maximum and minimum.
Here are some examples and how math is used in order to solve issues.
1-Maximizing the profit
In economic, profit maximization is the process by which a firm determines the price
and output level that returns the greatest profit. There are several approaches to this
problem. The total revenue -- total cost method relies on the fact that profit equals
revenue minus cost.
Ö How math is used in economic to get maximum profit??
Example 1
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
This example involves
maximizing profit when
the demand and averagecost functions are known.
profit occurs.
c- Determine the maximum profit.
In this example, we use the word monopolist. Under a situation of monopoly, there
is only one seller of a product for which there are no similar substitutes, and the
seller _that is, the monopolist _controls the market. By considering the demand
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equation for the product, the monopolist may set the price (or volume of output)so
that maximum profit will be obtained.
Solution:
We know that
profi= tR- tC
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
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= $17,420
Example 2
A business, in marketing a certain item, has discovered that the demand for the item is
represented by the equation
X=2500/p2
Assuming that the total revenue R is given by R =xp and the cost for producing x
items is given by C= 0.5x +500, find the price per unit that yields a maximum profit.
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Solution
1- let p represent the profit
2- Since we are seeking a maximum profit , the primary equation is
P=R –C
3- Using the given equation for R and C, we have
P=R –C =xp –(0.5x+ 500)
Solving for p in the equation x =2500/ p2, we have:
P= 50/ x
Thus the profit equation is
P= x (50/ x )- 0.5 -500 =50 x - 0.5x -500
4- Setting the derivative =0
p  = 25 / x -0.5=0
 x = 25/0.5 =50
 x=2500
Finally we conclude that the maximum profit occurs when the price is:
P=50/ 2500 =50/50= $1.00
Ö In order to maximize profit we look for the critical numbers of p ,that is ,the
number where the marginal profit is 0.But if
p   R ( x)  C ( x)  0
Then
Therefore:
R( x)  C ( x)
If the profit is a maximum, then
Marginal revenue =marginal cost
To ensure that this condition gives a maximum we could use the Second Derivative
test.
p ( x)  R ( x)  C ( x)  0
R ( x)  C ( x)
And this condition says that the rate of increase of marginal revenue is less than the
rate of increase of marginal cost. Thus, the profit will be a maximum when
R( x)  C ( x)
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and
R ( x)  C ( x)
Example 3
Determine he production level that will maximize the profit for a company with
cost and demand functions
C(x) =3800 +5x –(x2/1000)
p(x) =50- (x/100)
Solution:
The revenue function is
R(x)=xp= 50x –( x2/1000)
So the marginal revenue function is
And the marginal-cost function is
R( x)  50  ( x / 50)
C ( x)  5  ( x / 500)
Thus the marginal revenue is equal to marginal cost when
50  ( x / 50)  5  ( x / 500)
 x  2500
To check that this gives a maximum we compute the second derivative:
R( x)  1 / 50
C ( x)  1 / 500
Thus R( x)  C ( x) for all values of x .there for, a production level of 2500 units
will maximize the profit.
2-Minimizing the cost or the average 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 (1):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?
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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)
C¯=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 this is the critical number or q= -40 which is rejected because it is negative
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
 (d²c¯/dq)=800/q³
 (d²c¯/dq)(40)= 800/(40)³=0.0125>0
Which is positive for q=40. Thus, C¯ has a relative minimum when q=40.
We note that C¯ is continuous for q >0. 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
Level of out put will average cost per minimum is q=40
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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.
Solution: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
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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 (1):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.
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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
 -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?
Solution: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,
(dR/dq)= 80-2q=0
-2q=-80 we divided by -2
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Q=40
Thus, 40 is the only critical value. Now we see whether this gives a maximum.
Examining the first derivative for 0≤q<40, we have (dR/dq) >0, so r is increasing. If
q>40, then (dR/dq) <0, so r is decreasing. Because to the left of 40 we have r
increasing, and to the right r is decreasing, we conclude that q=40 gives the absolute
maximum revenue, namely,[80(40)-(40) ²]/4=400
4- Price elasticity of demand
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
n= (p/x)(dx/dp)
(Where n is the greek lowercase letter eta).For a given price,if lnl<1, the demand is
said to be inelastic; if lnl>1, the demand is said to be elastic
Example (1)
Show that the demand function , x= 2500/P^2, is elastic.
Solution
n= ( 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.
Ö Elasticity has an increasing relationship to the total revenue:
1. If the demand is elastic, then the decrease in the price per unit is accompanied by a
sufficient increase in unit sales to cause the total revenue to increase.
2. If the price is inelastic, then the increase in unit sales accompanying a decrease in
price per unit is not sufficient to increase revenue and the total revenue decrease
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R
elastic
(dR/dx)>0
inelastic
(dR/dx)<0
X
Revenue Curve
Example ( 2):
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 given 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
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Conclusion
Guide for solving applied max_man problems:
Step1. When appropriate, draw a diagram that reflects the information in the
Problem.
Step2. Set up a function for the quantity that you want to maximize or
Minimize.
Step3. Express the function in step2 as a function of one variable only, and note the
domain of this function. The domain may be implied by the nature of the problem
itself.
Step4. Find the critical values of the function. After testing each critical value,
determine which one gives the absolute extreme value you are seeking .if the domain
of the function includes endpoints, be sure to also examine function values at these
endpoints.
Step5. Based on the results of step 4, answer the question(s) posed in the
problem
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