Application Of Derivatives To Business And Economics Presented by: Fatma Al-Hassan Nada Abdulkareem Noor Al-Ghanim Tamadher Al-Thani Tahra Dina

Presentation content: Introduction to Application of derivatives. The demand function. The cost function. (Tamader) The revenue function. (Tahra) The profit function(& Maximizing it). (Dena) Examples 1.(Nada) Examples 2.(Noor)

The Demand Function.. and it computes via p(x)= p * x The demand function relates the price p(x) of a specific unit of a product to the units produced. and it computes via p(x)= p * x

A general Example: A drama takes place in a big theater that can take 10,000 people. At first the tickets cost QR50 , it was expected that around 5,000 people would come , and then when the tickets value were lowered to QR40 the audience attendance was expected to rise to 2000. Find the demand function assuming that this is a linear Function?

Solution First we find the equation of the straight line through the following points: (5000,50),(7000,40) Secondly we find the slope: m = (40-50)/(7000 - 5000) = -10/2000 = -1/200

So the equation of this line is: P(x) - 40 =(-1/200).(x - 7000) P(x)=-1/200.(x - 7000)+40 So the demand function is : P(x)= -1/200x+35+40 = -1/200x+75

The cost Function.. C(x) = p(x) * x It relates the total cost C(x) of producing an x number of units of a product to that number x C(x) = p(x) * x

Total cost example:- If a TV company wants to produce 10 TVs and each TV cost 8,000 QR, find total cost gain from selling this amount of TVs:- The solution:- C(x) = p(x) * x C(x) = (10*8000) * 8000 C(x) = 640,000,000 QR

The Revenue Function.. Revenue in economics means: Amount received or to be received from customers for sales of products or services.

Revenue Functions: IN general, businesses are concerned with total revenue as well as costs. Recall that if R(x) is the revenue received from the units sold as some commodity, then the derivative R'(x) is called the marginal revenue. the total revenue is compute via R(x) = x · p(x)

Example The demand equation of a certain product is p(x)=8-1/2x dollars. Find the revenue: R(x)= x.p(x) =x(8-1/2x) =8x-1/2x2

The Profit Function →x.p(x) – C(x) Once we know the cost function C(x) and the revenue function R(x), we can compute the profit function P(x) from P(x) = R(x) − C(x). →x.p(x) – C(x)

Notice that : A maximum profit is reached when: 1- The first derivative of P(x) when P´(x) is zero or doesn’t exist And 2- The second derivative of P(x) is always negative P″(x)<0

Another important notice is: Since P(x)= R(x)- C(x) when P´(x)=0 this is the critical point, and since P″(x)<0 the parabola will be concave downward Another important notice is: Since P(x)= R(x)- C(x) Then a maximum profit can be reached when: R'(x)-C'(x)=0 and R″ (x)-C″(x)<0

Example 1: A factory that produces i-pods computed its demand, and costs and came to realize that the formulas are as follows: p(x)= 100 - 0.01x C(x)= 50x + 10000

Find: 1-The number of units that should be produced for the factory to obtain maximum profit. 2-The price of the unit. R(x)= x . p(x) =x.(100 - 0.01x) = 100x - 0.01x2 Now: P(x)=R(x) - C(x) = 100x - 0.01x2 –(50x + 10000) = -0.01x2 + 50x - 10000

Now the maximum point in the parabola will occur When P'(x)=0 ; P"(x)<0 P(x) = -0.01x2 + 50x - 10000 P'(x)= -0.02x + 50 → -0.02x + 50 =0 → -0.02x = -50 → x= 2500 P'(2500)= -50 + 50=0 So the P'(X)=0 When x=2500

Now we will find the second derivative: P″(x)= -0.02 P″(2500)= -0.02<0 So x=2500 is at a local maximum

Graph

Finally to find the price that is needed to be charged per unit we return to the demand function: = 100-25= $75

Example 2: A factory director discovered that the total cost C(x) of producing x units of extremely sophisticated PC’s is: C(x) = 0.5x2 + 2000x + 24000000 And that the price p of the unit is subject to the demand equation: p(x) = -0.5x + 12000 Find The number of units that should be produced for the factory to obtain maximum profit. The maximum profit obtained.

Solutions We have: P(x) = R(x) – C(x) = x . p(x) – C(x) = x (-0.5x + 12000) – (0.5x2 + 2000x + 24000000) = - x2 + 10000x - 24000000

The maximum Profit P(x) = - x2 + 10000x – 24000000 Letting P′(x) = 0 , we get: x = 5000 Thus x = 5000 is a critical point We also have: P′′ (x) = - 2 → P′′ (5000) = - 2 < 0 Thus x=5000 is a point of local maximum

P(x) = - x2 + 10000x – 24000000 At x=5000, we have: The profit: P(5000) = - (5000)2 + 10000(5000) – 24000000 = ( - 25 + 50 - 24). 106 = ( 50 – 49 ). 106 = 106

Thank you