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Problem 1
The manufacturing cost of an item consists of Rs.6000 as over heads, material cost Rs. 5 per
unit and labour cost Rs.260xfor x units produced. Find how many units must beproduced so
that the average cost is minimum.
Problem 2
The demand function of a particular firm is given by: 5p + 2q – 25 =0. Your task is to find
the volume of production when the firm achieves maximum revenue.
Solution:
Given,
Demand Function 5p+2q-25=0
P=-2q/5 +5
Revenue function R(x)=pq = -2q2/5+5q
dR/dq = -4q/5 +5=0
4q/5 = 5;
q= 25/4 = 6.25
d2R/dq= -4/5 < 0
Therefore, Revenue is maximum at q = 6.25
Problem 3
A furniture manufacturing company developed its daily profit function as
P(x) =-0.007x2 +3x – 80 which is based on manufacture and sale of “x” units of furniture.
Your task is to find the number of furniture the company must manufacture and sell per day
to achieve maximum profit. You are also required to report the profit per furniture when
maximum profit is achieved.
. Given that,
Profit function, P(x) =-0.007x2 +3x -80
Revenue function, R(x) = px
For maximum profit,
dp/dx =0
=>d/dx (-0.007x^2 +3x -80) =0
=>x=214.2
Now d2p/dx2 =d/dx (dp/dx)
=d/dx (-0.014x +3)
= -0.014<0
Therefore given profit function is maximum at x=214.2 and maximum profit
=0.007*(214.2)2 +3*214.2 -80
=241.5
Therefore required number of furniture the company must manufacture and sell per day to
achieve maximum profit =214.2 units
Again,
Profit per furniture =maximum profit/no. Of furniture
=214.5/214.2
=1.12 rupees
Problem 4
The cost function of a departmental store related to a particular item is given by
TC(x)=0.045x3 +3x2+35x +20 where TC represents total cost and x is the number of
particular item the store is only selling. A tax at the rate of Rs.10.5 per unit of output is
imposed and adds it to its cost .if the market demand function is given by 2050-5x where Rs.
“p” is the price per unit of the particular item sold. How many such items the store should sell
to reach at the maximum profit and also find the maximum profit.
Solution. Given that,
Total cost function, TC(x) =0.045x3+3x2 +35x+20
Demand function=2050-5x
Let ‘R’ be the revenue then revenue function is given by,
R(x) = px
= (2050-5x) x
=2050x-5x^2
Cost function due to imposition of tax=0.045x^3+3x^2+35x+20+10.5x
Now profit function
P(x) =R(x)-C(x)
=2050x-5x^2-(0.045x3+3x2+35x+20+10.5x)
=-0.045x3-5x2-8x2+2004.5x-20
For maximum profit
dp/dx =0
=>d/dx (-0.045x3-8x2+2004.5x -2050)=0
=>0.135x2+16x-2004.5=0
=>x¿−16±√162−4∗0.135∗(−2004.50)
2∗0.135¿
¿
=>x=76.2,-194.7(impossible)
Now ,d2p/dx2 =d/dx(dp/dx)
=d/dx (-0.135x2-16x+2004.5)
At x=76.2, d2p/dx2 =-36.57<0
So profit is maximum when x=76.2 units and
Maximum profit, P =-0.045*(76.2)2 -8*(76.2)2 +2004.5 *(76.2) -20
=86361.2
Therefore required number of items the store should sell=76.2units.
Problem 5
With the help of a statistician and economist the cost accountant of a firm has designed the
profit function of a firm which is given by P(x) =0.0275x2+575x-1750, where x represents the
output of the firm. From experience it has also been found that output of the firm varies from
a minimum output level of 0 to maximum output level of 45. Your task is to find the value of
“x” which maximizes the total profit of the firm and also the amount of maximum profit at
such level of output.
Solution given profit function P(x) =0.02752+575x-1750
For maximum value of x,
dp/dx=d/dx (0.0275x2+575x-1750)
=- 0.0820x2+575
dp/dx =0
=>0.0825x2+575=0
=>x2=6969.69
=>x=83.48
Again, d2p/dx2=d/dx(dp/dx)
=d/dx (-0.0825x2+575)
=-0.165x
At x=83.48, d2p/dx2=-13.77<0
So, profit is maximum when x=83.48
And maximum profit p,
=0.0275*(83.48)3+575*(83.48)-1750
=15998.5+48001-1750
=62249.5 rupees
Problem 6
Suppose ‘p’ is the price of soft drinks when there is a sale of ‘q’ number of bottles sold. An
economist determined the relationship between p and q as given by: p= {250/ (4q+3)} -2.5.
As a management consultant you need to advice the company board regarding the marginal
revenue of selling 100 bottles of soft drinks.
Solution
Given that: p=price of bottle
q=no of bottles.
P= {250/ (4q+3)} -2.5
Now total revenue, R(x) =pq
= [{250/ (4q+3)} -2.5] q
=250/ (4q+3) q -2.5q
= {250q – 2.5(4q + 3)}/ (4q +3)
= (250q – 10q2 -7.5q)/ (4q +3)
Marginal revenue, MR = d/dq (TR)
=d/dq [{250q – 10q2 -7.5q}/(4q + 3 )]
=[(4q + 3)d/dq (250q – 10q2 – 7.5q) – (250q -10q2 – 7.5q) d/dq(4q +
3)] /(4q + 3)2
= [(4q + 3)(250 – 20q – 7.5) – 4(250 – 10q2 -7.5q)]/(4q + 3)2
= [1000q – 80q2 – 30q +750 – 60q -22.5 – 1000q +40q2 +30q]/( 4q
+ 3 )3
= [-40q2 – 60q + 727.5] /(4q + 3)2
Marginal revenue of selling 100 bottles,
MR10 = {-40* (100)2 -60 q+727.5}/ (4q + 3 )
Marginal revenue of selling 100 bottles,
MR100 = ( -40* (100)2 – 60* 100 + 72705) / (4*100 + 3 )2
=-2.49
Therefore marginal revenue is negative; the firm should produce less to maximize profit.
Problem 7
A company finds that for the 500 units that are produced and sold the profit is Rs 40 per unit.
The profit on each of the units beyond 500 is decreased by Rs 0.10 times the number of
additional units produced. What is the output level to maximize profit?
Soln: Let X be the number of additional units produced beyond
500 units.
Therefore, Total Profit, p= (500*40)+(40-0.10x)*x
=2000+40x-0.10x2 ……………….(1)
Differentiating (1) w.r.t. x, we get
dp/dx=40-0.20x
For maximum or minimum we should have dp/dx=0
i.e., 40-0.20x=0
=> x=40/0.20=4000/20=200
Now, d2p/dx2= -0.20, which is negative.
Since d2p/dx2 is negative, x=200 is a point of MAXIMA.
Therefore the required number of units that the manufacturer would produce to
maximize profits= (500+200) units=700 units.
Problem 8
A company makes X kg of material A and Y kg of material B where Y=2X-2X2-5. The profit
on material A is Rs 200 and on B it is Rs 100 per kg. How much of material A should be
produced per day to maximize profit?
SOLN: Given, Profit margin on material A=Rs.200
Profit margin on material B=Rs.100
Therefore , Total Profit, P=200x+100Y
=200X+100(2X-2X2-5)
=200X+200X-200X2-500
=400X-200X2-500
For maximum or minimum we should have dp/dx=0
=>400-400x=0
=>400(1-x)=0
=>x=1
Now, d2p/dx2=-400 which is negative.
Therefore the company should produce 1 kg of material A per day to maximize
profit.
ANS: I kg.
Problem 9
A rectangular tennis court of 1800 square meters is to be fenced with 2 types of materials.
The shorter sides are made with fence material costing Rs.100 per meter and the other sides
with fence material costing Rs.50 per meter. Find the dimensions of the court to minimize
cost.
Solution: Given,
Cost of material fencing the shorter sides = Rs. 100 /meter.
Cost of material fencing the other sides = Rs.50 /meter.
Area of the rectangular tennis court = 1800 square meters.
Let the dimensions of the tennis court be ‘a’ meters and ‘b’ meters.
It is required to find the dimensions of the court to minimize cost.
The cost of fencing the shorter sides is = 2a x 100
The cost of fencing the shorter sides is = 2b x 50
The total cost of fencing the whole tennis court is C = 2a x 100 + 2b x 50……….. (i)
Again, area is given by the formula
a x b = 1800
b = 1800/a………………. (ii)
Putting the value of (ii) in (i), we get
C (x) = 2a x 100 + 2 x (1800/a) x 50
In order to minimize cost, we need to check the following conditions,
Necessary condition
C(x) = 0
(200a + 180000/a) = 0
200 – 180000/a2 = 0
a2 = 180000/200
a2 = 900
a = 30
Sufficient condition
It is required to find the second order derivative of C(x)
ie. C’’(x)
= (200 - 180000/a2)
= 360000/a3
For a=30, we get C’’(30) = 360000/(30)3 = 13.33
So C’’ (30) > 0
The cost function will be get the minimum value when a=30 and b = 1800/30 = 60
Thus when the dimensions of the court are 30 metres and 60 metres the cost to fence it will
be minimized.
Problem 10
A company has established that the revenue function in dollars is R(x) = 2x3 +40x2 +8x and
the costfunction in dollars is C(x) = 3x3 + 19x2 + 80x − 800.Find the price per unit to
maximize the profit.
Solution: Given,
Revenue function in dollars is R(x) = 2x3 +40x2 +8x and the cost function in dollars is C(x) =
3x3 + 19x2 + 80x − 800.
It is required to find the price per unit to maximize profit.
Let p be the price per unit.
To maximize the profit, need the profit function
P(x) = R(x) – C(x)
P(x) = (2x3 + 40x2 + 8x) − (3x3 + 19x2 + 80x − 800)
= −x3 + 21x2 − 72x + 800
Once the equation of the profit function is found, use the first derivative to find ‘x’ which is
the necessary condition
P′(x) = d/dx P(x)
P′(x) = d/dx (−x3 + 21x2 − 72x + 800)
P′(x) = −3x2 + 42x − 72
Now we have
=> P′(x) = 0
=> −3x2 + 42x − 72 = 0
=> −3(x − 2) (x − 12) = 0
The values of x are: x = 2; x = 12
Sufficient condition
Use second derivative test:
P′′(x) = d/dx P′(x)
P′′(x) = d/dx (−3x2 + 42x − 72)
P′′(x) = −6x + 42
At x = 2
=> P′′(2) = -6x2+42
= 30 > 0
ie. Relative minimum
At x = 12
=> P′′(12) = -6 x 12 + 42
= -72 + 42
= -30 < 0
ie. Relative maximum
To maximize the profit, the number of units should be x = 12.
Now to find the price per unit we get,
p =R(x)/x
= (2x3 + 40x2 + 8x)/x
= 2x2 + 40x + 8
At x = 12, p(12) = 2(12)2 + 40(12) + 8 = Rs.776 per unit.
So the price per unit to maximize profit is Rs.776 per unit.
Problem 11
A producer of educational CDs is producing an instructional CD. The producer estimates that
it will cost Rs. 10000 to produce the CD and Rs. 3.25 per unit to copy and distribute the CD.
If the selling price of the CD is Rs. 4.50. How many CDs must the producer sell to break-
even?
Problem 12
A manufacturing company finds that the daily cost of producing x items of a product is given
by C(x)= 210x+7000. If each item is sold for Rs.350,find the minimum no. that must be
produced and sold daily to ensure no loss.
Problem 13
The demand function faced by a firm is p=500-0.2x and its cost function is C=25x+10000
(here p=price, x=output and C=cost). Find the output at which the profits of the firm are
maximum. Also find the price that it will charge?
Problem 14
The total profit y in Rupees of a company is given by Y = -x2/400 + 2x – 80. The sale of x
bottles is considered for the financial year 2011-12. Find how many bottles the company
must sell to achieve maximum profit.
Solution:-
Y = -x2/400 + 2x – 80
dydx
= −1400
* 2x + 2
= −1200
x + 2
Necessary condition
dydx
= 0
−1200
x + 2
x200
= 2
x= 400
Therefore the given profit is maximum at x = 400 & the maximum profit is – (400)2/400 + 2 *
400 – 80 = 320
Problem 15
If the total cost function of a firm is: C(x) = 1/3x3-5x2+30x+10, where C is the total cost and x
is the output and price under perfect competition is given as 6,find for what values of x the
profit will be maximized. Examine both first and second order conditions.
Solution:
Given C(x) = 1/3x3-5x2+30x+10 then marginal cost
MC = d/dx { C(x) }
MC = 1/3(3x2) – 5(2x) + 30
MC = x2 – 10x +30
Under perfect condition,
dp/dx = (dr/dx)-(dc/dx) = 0
6 = x2 – 10x +30
x = 4,6
The second order condition:
d2C/dx2 = 2x – 10
For x = 4
d2C/dx2 = -2(<0) this shows that for the output level, x= 4, the firm is in equilibrium
For x = 6
d2C/dx2 = 2(>0) this shows that for the output level, x= 6, the firm earns maximum profit,
given by:
P = 6x – 1/3x3 + 5x2 – 30x – 10
P = -46
The negative profit implies a loss. Since the loss is more than the fixed cost of Rs 10,
therefore it may be advised to stop production when P= Rs 6.
Problem 16
A firm knows that the demand function for one of its products is linear. It also knows that it
can sell 1000 units when the price is Rs.4 per unit and it can sell 1500 units when the price is
Rs.2 per unit. Determine :
(i) the demand function
(ii) the total revenue function
(iii) the average revenue function
(iv) the marginal revenue function.
Problem 17
The marginal revenue function of a commodity is given as MR =12-3x2 +4x . Find the total
revenue and the corresponding demand function.
Problem 18
A manufacture’s marginal revenue function is given byMR = 275-x-0.3x2 . Find the increase
in the manufacturer's total revenue if the production is increased from 10 to 20 units.
Ans: MR=275-x-.3x2
Revenue =∫ (275-x-.3x2)dx
=275x-x2/2-.3x3/3
=275x-x2/2- .1x3
R=275x-x2/2- .1x3
No when x=10
R1=2600
When x=20
R2=Rs4500/-
Increasing R=R2-R1
=Rs 1900/-
Problem 19
The cost function of producing x units of a product is given by C(x)=√(ax+b). Where a, b are
positive. Using derivatives show that the average and marginal cost curves fall continuously
with increasing output
Ans: c(x)=(ax+b)1/2
Avg cost=(ax+b)1/2/x
MC=d/dx(ax+b)1/2/x
=a/2(ax+b)3/2
Problem –20
A firm manufactures two products. The revenue function that describes total revenue from
sales of these products is given by
R = 8x + 5y +2xy – x2 – 2y2 + 20
Where, x is thousands of units of product one and y is thousands of units of product
two. Calculate the values of x and y that leads to maximum revenue. In case there is an
additional constraint x + y = 14, calculate the modified optimum values of x and y
Solution
The revenue function of the firm is
R = 8x + 5y + 2xy – x2 – 2y2 + 20
Or, R = 8x + 5(14 – x) + 2x(14 –x) – x2 – 2 (14 – x)2 +20
= 8x + 70 - 5x +28x – 2x2 – x2 – 392 + 56x -2x2 +20
= 87x – 5x2 – 302
We know that for the revenue to be maximum, the necessary condition is,
dRdx
= 87 - 10x = 0
So, x = 8710
and y = 14 – 8710
= 5310
ddx
{dRdx
} = - 10………………………… (The value is minimum)
Hence the point of inflexion x = 8710
carries the maximum value for the above revenue
function.
The modified optimum values of x and y are x = 8710
, y = 5310
Problem 21
The marginal revenue function for a product is given by MR=6/(x-3)2 Find the total revenue
function and the demand function.
Ans: R=∫6/(x-3)2 dx
=6/(x-3)[x2/2-3x]
=3x2/(x-3)-18x
Demand function=3x/(x-3)-18
Problem 22
The cost of manufacturing an item consists of Rs.3000 as over heads, material cost Rs. 8 per
item and the labour cost x2/30for x items produced. Find how many items must be produced
to have the average cost as minimum.
Soltn; Total cost
C(x)= 3000+ 8x+x2/30
AC=3000/ x+8+x2/30
Now,d(AC)/dx=-3000/x2+1/30
Therefore
d(AC)/dx=0
Þ -3000/x2+1/30=0,Þ x= 600
at x = 600
d2(AC)/dx2=+6000/x3>0,x=600
Hence AC is minimum when x = 600
23. For the cost function C = 2000 + 1800X – 75X2 + X3 , find when the total cost C is increasing
and when it is decreasing. Also discuss the behavior of the marginal cost (MC).
Solution:
Given,
Cost function C = 2000 + 1800X – 75X2 + X3
Necessary conditions:
dCdx
= 0
d (2000+1800 X –75 X2+X3)dx
= 0
1800 – 150 X + 3X2 = 0
X2 - 50X + 600 = 0
(X-20)(X-30) = 0
X = 20 or X = 30
Now,
i) 0 < X < 20, dCdx
> 0 e.g. when X = 10 then dCdx
= 600 >0
ii) 20 < X < 30, dCdx
<0 e.g. when X = 25 then dCdx
= - 75 < 0
iii) X > 30, dCdx
> 0 e.g. when X = 40 then dCdx
= 600 > 0
Therefore, C is increasing for 0 < X < 20 and X > 30
And, C is decreasing for 20 < X < 30
Again, we know that marginal cost,
MC = dCdx
= 1800 – 150 X + 3X2
And d (MC )dx
= - 150 + 6X
According to necessary condition,
d (MC )dx
= 0
- 150 + 6X = 0
X = 25
Now,
i) 0 < X < 25, d (MC )dx
< 0 e.g. X = 10 then d (MC )dx
= -90 < 0
ii) X > 25, d (MC )dx
> 0 e.g. X = 30 then d (MC )dx
= 30 > 0
Hence, MC is decreasing for X < 25 and increasing for X > 25
24. A firm produces X tones of output at total cost C = (1
10X3 – 5X2 +10X + 5). At what level of
output will the marginal cost and the average variable cost attain their respective minimum?
Solution:
Given,
Cost function, C(X) = (1
10X3 – 5X2 +10X + 5)
Variable Cost = 1
10X3 – 5X2 +10X
Let the Marginal Cost be MC
Let the Average Variable Cost be AVC
We know that, Marginal Cost, MC = d (C)dx
MC = 3
10X2 – 10X + 10
And,
Average Variable Cost, AVC = VariableCost
X
= 1
10X2 – 5X +10
i) Let us consider, Y = MC = 3
10X2 – 10X + 10
Differentiating Y with respect to X, we get
= dYdx
= 35
X – 10
Now, marginal cost is minimum when dYdx
= 0 and d2Yd X2 > 0
Necessary condition:
dYdx
= 0
35
X – 10 = 0
X = 503
Thus, when X = 503
, d2Yd X2 =
35
> 0 i. e. MC is minimum
Therefore, Marginal Cost attains its minimum at X = 503
units.
ii) Let Z = AVC = 1
10 X2 – 5X +10
Differentiating with respect to X, we get
dZdx
= 15
X – 5
AVC is minimum when dZdx
= 0 and d2Zd X2 >0
Therefore,
Necessary condition:
dZdx
= 0
15
X – 5 = 0
X = 25
Thus, when X = 25, d2Zd X2 =
15
> 0 i. e. AVC is minimum at X = 25 units
Therefore, Marginal Cost attains its minimum at X = 25 units.
Problem 25
Ans: x2=73.125
=>(x-1.5)(x-1)=73.125
=>x2-2.5x-71.625=0
Since cost can never be negative
Therefore 71.625+2x-x2=0
Now Dp/dx=2-2x
Therefore x=1
Now d2p/dx2=-2 so at x=1 the function is maximum.
At x=1 the printed functn
P=72.625
Problem 26
At a charity performance at the local cinema it is known that the attendance at a uniform
charge of Rs. P will be x=(a/p)-b, where a and b are constants. It is found that the cinema
which has 3,000 seats, is a half filled when Rs. 12 is charged but only 1/6 th of the seats are
empty when Rs. 9 is charged per seat. A) Find the constants a and b. B) Find the charge that
fills that show. Show that this charge provides the highest benefit to the charity assuming that
all proceeds go to charity.
Solution:
A)
The given equation is x=(a/b)-b
Total seats = 3000 = x
When seats are half filled the cost of ticket is Rs. 12, thus we get the equation,
X/2= (a/12)-b
1500 = (a/12)-b ---------------------------------------------------------(1)
When seats are 1/6th empty or 5/6th filled the cost of ticket is Rs. 9, thus we get the equation,
5X/6 = (a/9)-b
2500 = (a/9)-b ---------------------------------------------------------(2)
Now, Subtracting (1) from (2) we get
2500 – 1500 = {(a/9)-b} – {(a/12)-b}
1000 = (a/9) – (a/12)
1000 = a/36
a = 36000
Now substituting the value of a in equation (2) we get,
1500 = (36000/12) – b
1500 = 3000 – b
b = 1500
Therefore we get a = 36000 and b = 1500
B)
To get the highest benefit, there must not be any seats empty. Then only the optimum use of
resources will be there for maximum profit. Now we need to determine at what price the local
cinema will be completely filled up. So,
3000 = (a/p) –b
We already got the values of a = 36000 and b = 1500, replacing those values in the above
equation we get,
3000 = (36000/p) – 1500
36000/p = 4500
P = 36000/4500
p =8
Thus we get to know that at Rs. 8 the local cinema will be completely filled up and will be at
a profit maximum point and net revenue will be Rs.(8*3000) = Rs. 24000
Problem 27
A radio manufacture finds that he can sell x radios per week arRs. p each, where p-2(
100− x2
). His cost of production of x radios per week is Rs. (120+x2/2¿. show that his profit
is maximum when the production is 40 radios per week. Find also his maximum profit per
week.
Answer
Given,
Price, p = 2(100-x/2), for x radios
Cost, c = (120+x^2/2), for x radios
Therefore, Profit, P = 2(100-x/2).x - (120+x^2/2)
For maximum profit,
dP/dx = 0
x = 6230
For 40 radios
Price, p = 2(100-40/2) = Rs 160
Cost, c = (120+〖40〗^2/2) = Rs 920
For 40 radios
Profit, P = Total revenue – cost
= 160 x 40 – 920
= Rs 5480
Problem 28
A firm requires 50,000 units of material per annum. The cost of purchasing is Re. 0.5 per
unit, the cost of replishment of material is Rs. 28 and the cost of storing material is 21.6 per
cent per annum of the average rupee inventory. Find the optimum size and the corresponding
total cost.
Answer
C(x) = Ordering cost + Holding cost
= D/x . Cp + x/2 Ch
= 50000/x * 0.5 + x/2 * 1.6
For optimum size, the first order condition is
dC/dx = -50000/x2 x 0.5 + 1.6 = 0
or, -50000/x2 x 1.6 = ½
or, x2 = -50000 x 1.6 x 2
or, x = 160000
Also, d2C/dx2 – (2 x 50000 )/x3 x 1.6 > for x = 160000. Hence, economic lot size is x =
160000.
Problem 29
A firm produces x units of output per week ata a total cost of (13x3−x2+5x+3¿. Find the
output levels at which the marginal cost and the average cost attains their respective minima.
Answer
Given,
Total cost, TC = (13x3−x2+5x+3¿
Average cost (AC) = TC/x = (13x2−x1+5+3/ x
Marginal cost (MC) = d (TC)dx
= 3x2- 2x-+5
Now,
For MC to be minimum or maximum,
d (MC )dx
= 0
x = 13
and, d2MC/dx 2< 0
6<0
Hence MC is minimum at x= 6
Now,
For AC to be minimum or maximum,
d (AC )dx
= 0
x = 92
and, d2AC/dx 2< 0
3<0
Hence AC is minimum at x= 3
Problem 30
A wire of length 25 m is to be cut into pieces. One of the pieces is to be made into a square
and the other into a circle such that their combined area is minimum. Find the length of the
two pieces.
Answer
Let r be the radius of the circle whose perimeter is x. Then
2πr=x
r = x
2π
The perimeter of the square is 25 - x.
If A is the combined area then
A = πr2 + (25−x
4)2
= π ¿)2 + (25 – x)2/16
= x2/4π + (25 – x)2/16
dAdx
= x
2π –
(25−x)8
= 4 x−π (25−x )
8π
dAdx
= 0
4x – π (25−x) = 0
x = 25ππ+4
d2A/dp2 = 1
2π+
18
> 0 for all values of x
Therefore, A is minimum when x = 25ππ+4
Hence, the length of two pieces are
25ππ+4
and 25 - 25ππ+4
i.e., 100π+4
Problem 31
A manufacturer can sell x items per month at a price p = 300 – 2x rupees. Produced items
cost the manufacturer y rupees y = 2x + 1000. How much profit will yield maximum profits?
Answer
Profit (P) = Sale – total cost = x × p – y
= x (300 – 2x) – (2x + 1000) = 298x – 2x2 – 1000
For maximum profits , dpdx
= 0
i.e. ,298 - 4x = 0
or, x = 74.5
or, x = 74 (approx)
Again, d2p/dp2 = -4< 0
Hence, the profit will be maximum for 74 items
Problem 32
A hotel arranged a show on the following basis. The charges will be Rs5 if 40 or fewer
couple attended the show. If more than 40 couples attend the show, the charges for each
couple will be reduced by an amount equal to Re. 0.1 times the no of couples above 40. If the
hotel wants to maximizes revenue, how many it should admit?
Answer
Let x be the required no of couples. Then revenue from each couple be
5 - 1/10(x – 40) for x>= 40. Now total revenue,
R(x) = x {5 - 1/10(x – 40)}
= 5x - 1/10x2+40x = 45x - 1/10x2
For maximum revenue, we must have,
dR/dx = 45-2/10x = 0
or, x = 225
Also, d2R/dx2 = -2/10 < 0 for x = 225.
Thus R(x) is maximum for x = 225
33.The relationship between profit P and advertising cost X is given by P = 4000 X500+X - X. Find
X which maximizes P.
SOLUTION:
Given,
Profit P = 4000 X500+X – X
Differentiating with respect to X we get
dPdx
= (500+X ) 4000−(4000 X )(1)¿¿
- 1
= 2000000¿¿ - 1 --------------------- (i)
Profit is maximum when dPdx
= 0 and d2 Pd X2 < 0
Necessary Condition:
dPdx
= 0
2000000¿¿
- 1 = 0
1000 x √2 = 500 + X
1000 x 1.414 = 500 + X
X = 914
Differentiating (i) with respect to X we get
Sufficient Condition:
d2 Pd X2 = 4000000
¿¿
Therefore, when X = 914; d2 Pd X2 < 0
Hence profit is maximum at X = 914
34.The total cost and total revenue of a firm are given by C = X3 – 12X2 + 48X + 11 and R =
83X – 4X2 – 21. Find the output (i) when the revenue is maximum (ii) when the profit is
maximum.
SOLUTION:
Given
Cost function, C = X3 – 12X2 + 48X + 11
and Revenue function, R = 83X – 4X2 – 21
i) Differentiating R = 83X – 4X2 – 21 with respect to X
Necessary condition:
dRdx
= 83 – 8X
Sufficient Condition:d2 Rd X2 = -8
We know that,
Revenue is maximum when dRdx
= 0 and d2 Rd X2 < 0
Now,
dRdx
= 0
83 – 8X = 0
X = 838
Also,
d2 Rd X2 = -8 < 0 R is maximum.
When the output is X = 838
units, revenue is maximum.
35. A company producing dry cells introduces production bonus for its employees
which increases the cost of production. The daily cost of production C(x) for x numbers of
cells is ` (3.5x +12000)
a. If each cell is sold for ` 6, determine the number of cells that should be produced to ensure no
loss.
b. If the selling price is increased by 50 paise, what would be the breakeven point?
c. If at least 6000 cells can be sold daily, what price the company should charge per cell to
guarantee no loss?
Solution:
Given, Cost function, C (x) = 3.5x +12000
Revenue Function, R (x) = 6x
a. To ensure no profit and no loss,
C (x) = R (x)
Or, 3.5x + 12000 = 6x
Or, x = 4800
So, the company must produce 4800 items at the price of `6 so as not to incur
any loss.
b. New Revenue function R’(x) = 6.5 x
At Break-even point,
R’(x)-C(x) =0 (Zero)
Or, 6.5x – (3.5x+12000) = 0
Or, x = 4000
So, The Company, at the new selling price must product 4000 items so as not to incur any
loss.
c. Given, the amount of items produced daily, x =6000
Now Revenue function, R(x) =p 6000
For, no loss and no profit,
R(x) = C(x)
Or, 6000P = (3.5x + 12000)
Or, P = 5.5
Hence, we conclude that the company must sell the items at a price of `5.5 so as not to incur
loss or profit.
Where p is the selling price
Where x signifies the number of items, in this case it is 6000
4 3
PROBLEM 36
A hotel charges ` 80 per day for each room. However, special concession is available for each
room if more than 6 rooms are rented by each group; the rent of a room is decreased by ` 3 to
a minimum of ` 50. Each occupied room requires daily cleaning and other charges of ` 10.
The hotel also incurs a maintenance charge of ` 3 per room if it is not rented. The hotel has 50
rooms.
a. Compute the rent per room that a group has to pay if 50 rooms are rented by them.
b. Find the profit function of the number of rooms rented to a group in case only one group is
staying in the hotel at a time.
Solution:
a. Since the rent is reduced at the rate of`3 per room , the rent is reduced by
`(13 – 6)*3 = Rs.21 Thus the group has to pay at the rate of `(80-21) = `59 per room.
b. Let, P (x), R (x) and C(x) denote respectively the profit, revenue and cost function.
R(x) = 80x if 0 ≤ x ≤ 6.
Since, the rent is reduced to a minimum of `50 we get,
50 = 80 – 3 (x-6)
Or, x = 16
Or, R(x) = [80 – 3(x-6)]x, if 6 ≤ x ≤ 16
Or, R(x) = x (98-3x), if 6 ≤ x ≤ 16
Also, R(x) = 50 x, if x ≥ 16
Thus,
5 3
80x, if 0≤ x ≤ 6
R(x) = x (98 -3x) if 6 ≤ x ≤ 16
50x if x ≥ 16
Now, Let us compute the cost function C(x):
C(x) = 10x + 3 (50 –x) = 70 +150
[Each rented room requires a maintenance charge of `10 and the rooms which are not rented
requires Rs.3 for maintenance]
Since, P(x) = R(x) – C(x), We get
73x – 150, if 0 ≤ x ≤6
Or, P(x) = x(91 – 3x)*150 if 6 ≤ x ≤ 16
43x – 150 if x ≥ 16
6
PROBLEM 37
A company decides to set up a small production plant for manufacturing electronic clocks.
The total cost for initial set up is `9, 00,000. The additional cost for producing each clock
is`300. Each clock is sold at `750. During the first month 1,500 clocks are produced and sold:
a. Determine the cost function C(x) for the total cost of producing x clocks.
b. Determine the revenue function R(x) for the total revenue from sale of x clocks.
c. Determine the profit function P(x) for the profit from the sale of x clocks.
d. What profit or loss the company incurs during the first month when all the 1500 clocks are
sold?
e. Determine Break-even point.
Solution:
Given,
Fixed Cost = `9,00,000
Variable Cost = `300 per clock
a. Cost function, C(x) = Fixed cost +Variable cost
= 9,00,000 + 300x
b. Revenue Function, R(x) = (Selling Price) * (Quantity Sold)
= 750 x
Where x is the quantity sold.
7
c. Profit function, P(x) = R (x) – C(x)
= 750x – (9,00,000 + 300x)
= 450x – 9,00,000.
38.The total curve for the consumption of electricity is a linear function of the number of
units consumed. The cost per unit on the first 60 units is lower and on the balance over 60, is
higher. If the total cost for 100 units and the 150 units consumed are ` 62 and ` 102
respectively. Find out the equation of the total cost line for the consumption not less than 60
units and the charge for 180 units.
Solution:
According to question, the total cost is a curve of linear function
Hence, C(x) = mx + c…………………..(a)
When cost is `62 the number of units that are used is 100.
Therefore the equation (a), transforms,
62 = 100m + c……………………….(i)
When cost is `102 the numbers of units that are used are 152.
Therefore the equation (a) transforms,
102 = 150m + c………………………….(ii)
(ii) – (i), we get,
40 = 50m
Or, m = 0.8……………………………. (b)
Putting the value of m= 0.8 in equation (i),
We get,62 = 100*0.8 + c
Or, c = -18…………………..(c)
We can now write the values of m and c from equations (b) and (c) respectively to (a)
C(x) = 0.8 x – 18…………………… (M)
Now to find out the charges for 180 units we put x = 180 in the equation (M)
C(180) = 0.8 *180 – 18
Or, C(180) = 126, Hence we get that the charges for 180 units is `126.
Problem 39
A firm produces x units of output per week at a total cost of Rs. ( 13x3x2 + 5x + 3). Find the
level at which the marginal cost and the average variable cost attain their respective
minimum.
Answer
Given,
Total cost, TC = (13x3−x2+5x+3¿
Average cost (AC) = TC/x = (13x2−x1+5+3/ x
Marginal cost (MC) = d (TC)dx
= 3x2- 2x-+5
Now,
For MC to be minimum or maximum,
d (MC )dx
= 0
x = 13
and, d2MC/dx 2< 0
6<0
Hence MC is minimum at x= 6
Now,
For AC to be minimum or maximum,
d (AC )dx
= 0
x = 92
and, d2AC/dx 2< 0
3<0
Hence AC is minimum at x= 3
40.A firm wants to take a decision for introducing a new product. The cost of introducing the
product is estimated at `30,000. The variable cost per unit is `30 and the proposed selling
price is `50.
a. Find out the break-even level of production
b. Determine the profit on sales of 2,500 Units
Represent the cost, revenue and profit functions graphically.
Solution:
Fixed cost =`30,000
Variable cost per unit = `30
Selling price = `50
Cost function, C(x) = Fixed cost +variable cost
Or, C(x) = 30,000 + 30x
Revenue Function, R(x) = 50x
At break-even point,
C(x) = R(x)
Or, 30,000 + 30x = 50x
Or, x = 1500
Hence, we found that to reach the break-even level of production the firm must produce 1500
units.
Profit = R (x) - C(x)
Or, Profit = 50x – (30,000 + 30x)
Or, Profit = 20x – 30,000
Or, Profit = 20,000
Where x is the items produced
Given, x = 2500,
41)A vegetable seller profit function is P(x)= -2x²+4x+1. Find the amount that he will earn at
maximum profit?
Soln: Given, P(x)=2x²-4x+1
We check whether the given function is maximum or not
a)Necessary condition b)Sufficient condition
dP(x)/dx=0 =>d²p(x)/dx²<0 (to be maximum at
x=1)
=>d(-2x²+4x+1)/dx=0 =>-4<0
=>-4x+4=0
=>x=1
Thus the required condition is proved.
Therefore the maximum amount of profit the vegetable seller will earn is at x=1 is Rs
3
42)The cost function of a company is given as C(x)=5x³-3x and the revenue function is
R(x)=5x²+22. Find the profit function and check its maximum and minimum.
Soln: Given, C(x)=5x³-3x , R(x)=3x²+22
The profit function is given by P(x)= R(x)-C(x)
=5x²+22-5x³+3x
Now to check whether the given function is maximum or minimum
a)Necessary condition b)Sufficient condition
dP(x)/dx=0 d²p(x)/dx² (at x= -1)
=>d(5x²+22-5x³+3x)/dx=0 =>-30x+10
=>-15x²+10x +25=0 =>40>0 (minimum at x= -1)
=>-3x²+2x+5=0 now => -38<0 (maximm at
x=1.6)
=>x=-1,1.6
Hence the P(x) is maximum at x=1.6 and minimum at x= -1 .
43)The total revenue function of a bicycle company is given by R(x)=3x³-12x²+4. Find the
number of unit(bicycle) to be sold to get minimum average revenue where ‘x’ represents the
number of bicycle sold.
Soln: Given, R(x)=3x³-12x²+4x
The average revenue is given by AR(x)= R(x)/x
=(3x³+2x²+4)/x
=3x²-12x+4
We need to check the AR is maximum or minimum
a)Necessary condition b) Sufficient condition
dP(x)/dx=0 d²p(x)/dx² (at x= 2)
=>d(3x²+2x+4)/dx=0 =>6>0 (minimum)
=>6x-12=0
=>x =2
Thus the required condition is proved.
So the number of bicycle that is to be sold to get minimum average revenue is 2.
44.Find the maximum or minimum values of:
X^3-9X^2+15X-1
Solution: Let,
Y= X^3-9X^2+15X-1
Necessary Condition
Dy/dx=0
Differentiating with respect to x we have-
D(X^3-9X^2+15X-1)/dx= 0
=>3X^2-18X+15= 0
=>3(X^2-6X+5)=0
=>(X-5)(X-1)=0
=>X= 5,1
Sufficient Condition
Differentiating again with respect to x we have-
D^2y/dX^2= d(3X^2-18X+15)= 6X-18
Putting X=5,
D^2y/dX^2=6*5-18
=30-18= 12>0
Value is minimum at x=5.
Minimum value= 5^3-9*5^2+15*5-1
=-26
Again, Putting X=1,
D^2y/dX^2=6*1-18=6-18=-12<0
Value is maximum at x=1.
Maximum value= 1^3-9*1^2+15*1-1
=1-9+15-1
=6
Question 45:
Given the price in rupees per unit p = −3x2 + 600x,
Find :( a) the marginal revenue at x = 300 units.
(b) The marginal revenue at x = 100 units.
Solution: Given, The price in rupees per unit p= -3x2+600x where ‘p’ is the price.
(a) It is required to find the marginal revenue at x=300 units
Revenue function for producing x units is R(x) = p. x
= (−3x2 + 600x). x
= −3x3 + 600x2
Marginal revenue:
R′(x) = dR(x)/dx
= d/dx(-3x3+600x2)
= −9x2 + 1200x
Marginal revenue at x = 300
=>R′ (300) =dR(x)/dx at x=300
= −9 (300)2 + 1200(300) = −450 000
Interpretation: If production increases from 300 to 301 units, the revenue decreases by
Rs.450 000.
(b) It is required to find the marginal revenue at x=100 units
Revenue function: R(x) = p. x
= (−3x2 + 600x) . x
= −3x3 + 600x2
Marginal revenue: R′(x) =dR(x)/dx
= d/dx(-3x3+600x2)
= −9x2 + 1200x
Marginal revenue at x = 100
=> R′(100) =dR(x)/dx at x=100
= −9(100)2 + 1200(100) = 30 000
Interpretation: If production increases from 100 to 101 units, the revenue increases by
Rs.30000.
Question 46:
Given the revenue function in rupees R(x) = −5x3 + 800x2 and the cost function in rupees
C(x) = 357x2 + 1800x; find:
(a) The marginal profit at x = 10 units.
(b) The marginal profit at x = 100 units.
Solution: Given,
Revenue function R(x) = −5x3 + 800x2 and cost function C(x) = 357x2 +1800x
(a) It is required to find the marginal profit at x=10 units
We can calculate the profit function as
Profit function = Revenue function – Cost function
P(x) = (−5x3 + 800x2) − (357x2 + 1800x)
= −5x3 + 443x2 − 1800x
So marginal profit: P′(x) =dP/dx
= −15x2 + 886x − 1800
Marginal profit at x = 10
=> P′(10) =dP/dx at x=10
= −15(10)2 + 886(10) – 1800
= 5560
Interpretation: If production increases from 10 to 11 units, the profit increases by Rs.5560
(b) It is required to find the marginal profit at x=100 units
The profit function can be calculated as
Profit function = revenue − cost
P(x) = (−5x3 + 800x2) − (357x2 + 1800x)
= −5x3 + 443x2 − 1800x
So marginal profit: P′(x) =dP/dx
= −15x2 + 886x − 1800
Marginal profit at x = 10
=> P′(100) =dP/dx at x=100
= −15(100)2 + 886(100) − 1800
= −63200
Interpretation: If production increases from 100 to 101 units, the profit decreases by
Rs.63200
Question 47:
A rectangular tennis court of 1800 square meters is to be fenced with 2 types of materials.
The shorter sides are made with fence material costing Rs.100 per meter and the other sides
with fence material costing Rs.50 per meter. Find the dimensions of the court to minimize
cost.
Solution: Given,
Cost of material fencing the shorter sides = Rs. 100 /meter.
Cost of material fencing the other sides = Rs.50 /meter.
Area of the rectangular tennis court = 1800 square meters.
Let the dimensions of the tennis court be ‘a’ meters and ‘b’ meters.
It is required to find the dimensions of the court to minimize cost.
The cost of fencing the shorter sides is = 2a x 100
The cost of fencing the shorter sides is = 2b x 50
The total cost of fencing the whole tennis court is C = 2a x 100 + 2b x 50……….. (i)
Again, area is given by the formula
a x b = 1800
b = 1800/a………………. (ii)
Putting the value of (ii) in (i), we get
C (x) = 2a x 100 + 2 x (1800/a) x 50
In order to minimize cost, we need to check the following conditions,
Necessary condition
C(x) = 0
(200a + 180000/a) = 0
200 – 180000/a2 = 0
a2 = 180000/200
a2 = 900
a = 30
Sufficient condition
It is required to find the second order derivative of C(x)
ie. C’’(x)
= (200 - 180000/a2)
= 360000/a3
For a=30, we get C’’(30) = 360000/(30)3 = 13.33
So C’’ (30) > 0
The cost function will be get the minimum value when a=30 and b = 1800/30 = 60
Thus when the dimensions of the court are 30 metres and 60 metres the cost to fence it will
be minimized.
Question 48:
A company has established that the revenue function in dollars is R(x) = 2x3 +40x2 +8x and
the cost function in dollars is C(x) = 3x3 + 19x2 + 80x − 800. Find the price per unit to
maximize the profit.
Solution: Given,
Revenue function in dollars is R(x) = 2x3 +40x2 +8x and the cost function in dollars is C(x) =
3x3 + 19x2 + 80x − 800.
It is required to find the price per unit to maximize profit.
Let p be the price per unit.
To maximize the profit, need the profit function
P(x) = R(x) – C(x)
P(x) = (2x3 + 40x2 + 8x) − (3x3 + 19x2 + 80x − 800)
= −x3 + 21x2 − 72x + 800
Once the equation of the profit function is found, use the first derivative to find ‘x’ which is
the necessary condition
P′(x) = d/dx P(x)
P′(x) = d/dx (−x3 + 21x2 − 72x + 800)
P′(x) = −3x2 + 42x − 72
Now we have
=> P′(x) = 0
=> −3x2 + 42x − 72 = 0
=> −3(x − 2) (x − 12) = 0
The values of x are: x = 2; x = 12
Sufficient condition
Use second derivative test:
P′′(x) = d/dx P′(x)
P′′(x) = d/dx (−3x2 + 42x − 72)
P′′(x) = −6x + 42
At x = 2
=> P′′(2) = -6x2+42
= 30 > 0
ie. Relative minimum
At x = 12
=> P′′(12) = -6 x 12 + 42
= -72 + 42
= -30 < 0
ie. Relative maximum
To maximize the profit, the number of units should be x = 12.
Now to find the price per unit we get,
p =R(x)/x
= (2x3 + 40x2 + 8x)/x
= 2x2 + 40x + 8
At x = 12, p(12) = 2(12)2 + 40(12) + 8 = Rs.776 per unit.
So the price per unit to maximize profit is Rs.776 per unit.
Question 49:
Given the average cost in dollar per unit C = 357 x + 1800, find the marginal cost at x = 50
units.
Solution:
Given, average cost in dollar per unit C = 357 x + 1800
It is required to find the marginal cost at x=50
We can calculate the cost function as
Cost function: C(x) = C. x
= (357x + 1800). x
= 357x2 + 1800x
So Marginal cost: C′(x) =dC/dx
= 714x + 1800
Marginal cost at x = 50
=>C′(50) = dC(x)/dx at x=50
= 714(50) + 1800
= 37 500
Interpretation: If production increases from 50 to 51 units, the cost increases by 37500
dollars.
Problem50
The total cost and total revenue of a firm are given by C = x3 12x2 + 48x + 11 and R =
83x 4x2 21. Find the output
(i) when the revenue is maximum
(ii) (ii) when profit is maximum.
Given
Cost function, C = X3 – 12X2 + 48X + 11
and Revenue function, R = 83X – 4X2 – 21
ii) Differentiating R = 83X – 4X2 – 21 with respect to X
Necessary condition:
dRdx
= 83 – 8X
Sufficient Condition:d2 Rd X2 = -8
We know that,
Revenue is maximum when dRdx
= 0 and d2 Rd X2 < 0
Now,
dRdx
= 0
83 – 8X = 0
X = 838
Also,
d2 Rd X2 = -8 < 0 R is maximum.
When the output is X = 838
units, revenue is maximum.
iii) Let the profit function be P.
We know that,
Profit = Revenue – Cost
P = (83X – 4X2 – 21) – (X3 – 12X2 + 48X + 11)
= - X3 + 8X2 + 35X – 32
Differentiating with respect to X,
Necessary condition:
dPdx
= - 3X2 + 16X +35
Sufficient Condition: d2 Pd X2 = -6X + 16
Profit is maximum when dPdx
= 0 and d2 Pd X2 < 0
Equating the first derivative of Profit function P with 0, we get
dPdx
= 0
- 3X2 + 16X +35 = 0
3X2 - 16X – 35 = 0
(3X + 5 ) (X-7) = 0
X = −53
or X = 7
When X = −53,d2 Pd X2=−6(
−53
¿ + 16 = 26 > 0 P is minimum
When X = 7, d2 Pd X2 =−6(7¿ + 16 = - 26 < 0 P is maximum
Therefore, when X = 7 units, the profit is maximum.
