UGBS 202 – 2014/2015
Lecture Note 3
This lecture note was originally written by a student and partly modified and certified by the
instructor.
Extreme-Value Theorem
If a function 𝑓 is continuous on a closed interval [𝑎, 𝑏], then the function has both a maximum value and a minimum value on that interval [𝑎, 𝑏].
Note that one should guarantee that:
a. the function is continuous
b. the function is being analyzed in a closed interval
The symbol [ 𝑎, 𝑏 ] implies a closed interval and includes the endpoints 𝑎 and 𝑏 whereas (𝑎, 𝑏) implies an open interval and excludes the endpoints 𝑎 and 𝑏.
Procedure to find absolute extrema for a function 𝑓 that is continuous on [𝑎, 𝑏]
Steps:
1. Find the critical (relative extrema) values of 𝑓.
2. Evaluate the 𝑓(𝑥) at the end points 𝑎 and 𝑏 and the critical values in (𝑎, 𝑏).
3. The maximum value of 𝑓 is the greatest of the values found in the step 2 and the minimum value of 𝑓 is the least of the values found in step 2
Example:
Given 𝑝 =80−𝑞4 0 ≤ 𝑞 ≤ 80,
where 𝑞 = no. of units, 𝑝 = price per unit. At what value of 𝑞 will maximum revenue be attained? What is this maximum revenue?
Solution:
Revenue = price * quantity
(80 − 𝑞
4 )𝑞 = 20𝑞 − 1 4𝑞2
At the quantity where maximum revenue is attained, marginal revenue = 0. That is,
𝑑𝑟
𝑑𝑞= 𝑟′(𝑞) = 20 − 1 2𝑞 = 0
𝑞 = 40
𝑟′′(𝑞) < 0
Therefore 𝑞 = 40 is a relative maximum revenue value
However, since the analysis is being carried over a closed interval, we need to verify the endpoints to be sure that 𝑞 = 40 indeed yields the absolute maximum revenue.
At 𝑞 = 0, 𝑞 = 40 and at 𝑞 = 80
𝑟(0) = 0
𝑟(40) = 20(40) −1
4(40)2= 400
𝑟(80) = 20(80) −1
4(80)2= 0
Thus, the maximum revenue is attained at 𝑞 = 40.
APPLICATIONS OF RELATIVE MAXIMUM AND MINIMUM Illustration 1
Minimizing Average Cost
Suppose 𝐶 is the total cost of producing 𝑞 units of a product where, 𝑐(𝑞) =𝑞42+ 3𝑞 + 400. - At what level of output 𝑞 will average cost per unit be minimum?
- What is this minimum value? Solution
Let 𝐴𝐶 and 𝐶 be the average cost and total cost resp.
𝐴𝐶 =𝐶 𝑞
⇒
𝑞2
4 + 3𝑞 + 400
𝑞
⇒𝑞 4+ 3 +
400 𝑞
- At the quantity where average cost is minimum, 𝐴𝐶′(𝑄) = 0
𝐴𝐶′(𝑄) =1
4− 400𝑞−2
⇒14−400𝑞2
⇒ 𝑞2= 1600 ⇒ 𝑞 = ±40
⇒ 𝑞 = 40 (since can’t have negative 𝑞)
To confirm whether 𝐴𝐶 is minimum at 𝑞 = 40, the 2nd derivative test is applied.
- 𝐴𝐶′′𝑄 = 800𝑞−3
⇒ 800𝑞−3> 0
Therefore minimum average cost occurs at 𝑞 = 40
The value of the minimum average cost at 𝑞 = 40 is then:
𝐴𝐶 =40 4 + 3 +
400 40 = 23
Illustration 2
Maximizing TV Cable Company Revenue
The Vista TV Cable Company currently has 100,000 subscribers who each are paying a monthly rate at GH¢40.00. A survey conducted reveals that there will be 1000 more subscribers for each GH¢0.25 decrease in rate. At what rate will maximum revenue be achieved and how many subscribers will there be at this rate?
Solution:
Revenue = Price × Quantity or Revenue = Rate × Number of subscribers
Let 𝑥 be the number of times rate is decreased. As an example, if we decrease the rate by 0.25, followed by 0.25, and followed by another 0.25 so that the rate is (40-3*0.2), then 𝑥 = 3.
∴ Rate =40 − 0.25𝑥
Number of subscribers = 100000 + 1000𝑥
So at any 𝑥, revenue =(40 − 0.25𝑥)(100000 + 1000𝑥)
= 4000000 + 15000𝑥 − 250𝑥2
At the maximum revenue, marginal revenue = 0
𝑑𝑟
𝑑𝑥= 𝑟′(𝑥) = 15000 − 500𝑥
⇒ 15000 − 500𝑥 = 0 ⇒ 15000 = 500𝑥 ⇒ 𝑥 = 30
From the 2nd derivative test, 𝑟′′(𝑥) < 0. Thus, at 𝑥 = 30, a ‘relative’ maximum revenue is attained. However, note that 0 ≤ 𝑥 ≤ 160. That is, the TV company can choose to set 𝑥 = 0 (meaning, no reduction) or set 𝑥 = 160 (i.e. charge nothing). We thus have to test the revenue particularly at 𝑥 = 0 since it is obvious revenue is zero at 𝑥 = 160
𝑟(0) = 4𝑀 and 𝑟(30) = 4𝑀 + 15000(30) − 250(30)2= 4.225𝑀 > 4𝑀 Thus maximum revenue is attained when the current rate is reduced 30 times. Therefore maximum revenue = 4000000 + 15000(30) − 250(30)2
More Illustrations
Bani Hostels has available 170 rooms. When the rent is GH¢2,100, all rooms will be rented out. However, there will be 4 unoccupied rooms for every GH¢40.00 increase in the rent charged. If the cost of running the hostel is a fixed cost of GH¢100,000 and a variable cost of GH¢880.00 per an occupied room, what is the rent charge that will maximize Bani Hostel’s profit and what will be the number of rooms that should be made available for rent?
Suppose 𝑛 is the number of recipient of a healthcare benefit, after 𝑡 years, where 𝑛 is in
thousands of elderly people and given as 𝑛 =𝑡33− 6𝑡2+ 32𝑡, 0 ≤ 𝑡 ≤ 12. At what values of 𝑡 will the number of elderly receiving benefits start to fall?
Solution:
𝑑𝑛
𝑑𝑡 = 𝑡2− 12𝑡 + 32 𝑡2− 12𝑡 + 32 = 0
𝑡2− 4𝑡 − 8𝑡 + 32 = 0
𝑡(𝑡 − 4) − 8(𝑡 − 4) = 0 (𝑡 − 8)(𝑡 − 4) = 0 𝑡 = 8, 𝑡 = 4
Note: You could also use the quadratic equation formula, say 𝑡 =−𝑏±√𝑏2𝑎2−4𝑎𝑐 to find 𝑡.
INTEGRATION
DifferentialsGiven 𝑦 = 𝑓(𝑥), the derivative is given as:
𝑑𝑦
𝑑𝑥 = 𝑓′(𝑥) 𝑑𝑦 = 𝑓′(𝑥)𝑑𝑥
𝑑𝑦 is the differential of 𝑦, (that is, the infinitesimal change in 𝑦 due to an infinitesimal change in 𝑥). Integration says that if we were to add up all the 𝑑𝑦, the result should be equal to 𝑦. That is
𝑦 = ∫ 𝑑𝑦 = ∫ 𝑓′(𝑥)𝑑𝑥
Where the symbol ∫ is called the integral sign. Let 𝑓′(𝑥) = 𝐹(𝑥)
𝑦 = ∫ 𝑑𝑦 = ∫ 𝐹(𝑥)𝑑𝑥
The Indefinite Integral
Example 1
Let 𝑓(𝑥) = 2𝑥 and 𝐹(𝑥) = 𝑥2
𝐹′(𝑥) = 2𝑥 = 𝑓(𝑥)
Then we can conclude that 𝐹(𝑥) = 𝑥2 is the anti-derivative of 𝑓(𝑥) = 2𝑥. Intuitively, if we were to perform the opposite of differentiation on 𝑓(𝑥), we will get 𝐹(𝑥).
Example 2
Let 𝑓(𝑥) = 2𝑥 and 𝐹(𝑥) = 𝑥2+ 5
𝐹′(𝑥) = 2𝑥 = 𝑓(𝑥)
Thus, 𝐹(𝑥) = 𝑥2+ 5 is the anti-derivative of 𝑓(𝑥) = 2𝑥 Example 3
Let 𝑓(𝑥) = 2𝑥 and 𝐹 (𝑥) = 𝑥2− 4
𝐹′(𝑥) = 2𝑥 = 𝑓(𝑥)
Again, 𝐹(𝑥) = 𝑥2− 5 is the anti-derivative of 𝑓(𝑥) = 2𝑥.
Examples 1, 2 and 3 implies that 𝑓(𝑥) = 2𝑥 could have many anti-derivatives. However, 𝑓(𝑥) must have exactly one anti-derivative. To overcome the problem of having 𝑓(𝑥) having multiple anti-derivatives, we will add a constant 𝑐 to the anti-derivative 𝐹(𝑥).
Integration
In essence we have 𝑦 = ∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝑐
Elementary Integration Formulas
1. ∫ 𝑘𝑑𝑥 = 𝑘𝑥 + 𝑐, where 𝑘 and 𝑐 are constants E.g.∫ 2𝑑𝑥 = 2𝑥 + 𝑐
2. ∫ 𝑥𝑎𝑑𝑥 =𝑥𝑎+1
𝑎+1+ 𝑐
E.g.∫ 𝑥3𝑑𝑥 =𝑥3+13+1+ 𝑐 =𝑥44+ 𝑐
3. ∫ 𝑥−1𝑑𝑥 = ∫1𝑥𝑑𝑥 = 𝐼𝑛𝑥 + 𝑐 (remember 𝑑𝑥𝑑 (𝐼𝑛𝑥)?)
4. ∫ 𝑒𝑥𝑑𝑥 = 𝑒𝑥+ 𝑐
5. ∫ 𝑘𝑓(𝑥)𝑑𝑥 = 𝑘 ∫ 𝑓(𝑥)𝑑𝑥, where 𝑘 is a constant E.g.∫ 2𝑥2𝑑𝑥 = 2 ∫ 𝑥2𝑑𝑥 = 2𝑥33+ 𝑐
6. ∫[𝑓(𝑥) ± 𝑔(𝑥)] 𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 ± ∫ 𝑔(𝑥)𝑑𝑥
E.g.∫(𝑥2− 2𝑥)𝑑𝑥 = ∫ 𝑥2𝑑𝑥 − ∫ 2𝑥𝑑𝑥
7. Find the integral ∫(𝑥2+ 2𝑥)𝑑𝑥
Solution:
𝑦 =𝑥3
3 + 2𝑥 + 𝑐
Integration with Initial Conditions
To obtain the actual value of 𝑐 in the integration formula will require some initial knowledge (or condition) about the function under study.
Illustration 1
Suppose 𝑓(𝑥) = 𝑥2+ 𝑐 and that at 𝑥 = 1, 𝑓(1) = 4. Then
𝑓(1) ⇒ 12+ 𝑐 = 4
⇒ 𝑐 = 3
Thus, the actual function is 𝑓(𝑥) = 𝑥2+ 3
Illustration 2
If 𝑦 is a function of 𝑥 such that 𝑦′ = 𝑑𝑦/𝑑𝑥 = 8𝑥 − 4 and 𝑦(2) = 5, find 𝑦.
∫ 𝑑𝑦 = ∫(8𝑥 − 4)𝑑𝑥 𝑦 = 4𝑥2− 4𝑥 + 𝑐
𝑦(2) ⇒ 4(2)2− 4(2) + 𝑐 = 5
⇒ 16 − 8 + 𝑐 = 5 ⇒ 𝑐 = 5 − 8 ⇒ 𝑐 = −3 ∴ 𝑦 = 4𝑥2− 4𝑥 − 3
Illustration 3 Practical Question:
Finding the demand function from Marginal Revenue
The Marginal Revenue function for a manufacturer’s product is 𝑑𝑞𝑑𝑟= 2000 − 20𝑞 − 3𝑞2. Find the unit price 𝑝(termed the function demand).
Solution:
𝑟(𝑞) = ∫(2000 − 20𝑞 − 3𝑞2)𝑑𝑞 = 2000𝑞 − 10𝑞2− 𝑞3+ 𝑐
Although we are not given any initial condition, note that in practice when 𝑞 = 0, 𝑟(0) = 0. Thus, 𝑐 = 0, and 𝑟(𝑞) = 2000𝑞 − 10𝑞2− 𝑞3
Price = revenue /quantity
𝑝 =𝑟(𝑞)
Illustration 4
Given weekly fixed cost as Ghc 4000, find the weekly total cost when the marginal cost function is:
𝑑𝑐
𝑑𝑞= 0.000001(0.002𝑞2− 25𝑞) + 0.2 , where 𝐶 is total cost of producing 𝑞 units of products per week.
What is the cost of producing 𝑞 = 10,000? Solution
𝐶(𝑞) = ∫(0.000001(0.002𝑞2− 25𝑞) + 0.2)𝑑𝑞 = 0.000001(0.002𝑞3
3 −
25𝑞2
2 + 0.2𝑞 + 𝑐1
In this case, when 𝑞 = 0, 𝑐 = 4000. That is, the fixed cost is incurred whether production occurs or not.
⇒ 𝑐1= 4000 and
𝐶(𝑞) = 0.000001(0.002 3 𝑞3−
25
2 𝑞2+ 0.2𝑞 + 4000
At 𝑞 = 10,000
