• 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