Application of derivatives to Business and economics

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Application of derivatives to Business and
economics

• Presented by
• Amal Al-Kuwari
• Bashayer
• Noof
• Hind Nader

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Presentation content

• Introduction to Application of derivatives and
  its importance in the Business field
• The demand function
• The cost function
• The revenue function
• The profit function
• Examples

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The demand function

• The demand function is the function that relates
  the price p(x) of a specific unit of a product to
  the number of units x produced.
• p(x) is also called the demand function.
• p(x) p x

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The cost Function

• The cost function is the function that 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

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A general Example

• A football match takes place in a stadium that
  can take 60,000 people. At first the tickets cost
  QR30 , it was expected that around 30,000 people
  would come , then when the tickets value were
  lowered to QR20 the audience attendance was
  expected to rise to 5000.
• Find the demand function assuming that this is a
  linear Function?

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Solution

• First we find the equation of the straight line
  through the following points
• (30000,30) ,(35000,20)
• Secondly we find the slope
• m (20-30)/(35,000 - 30,000)
• -10/5000
• -1/500

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• So the equation of this line is
• P(x) - 20 (-1/500).(x - 35,000)
• P(x)-1/500.(x - 35,000)20
• So the demand function is
• P(x) -1/500x7020
• -1/500x90

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• Total cost example-
• If BMW company wants to produce 10 cars and each
  car cost 200,000 QR, find total cost gain from
  selling this amount of cars-
• The solution-
• C(x) P(x) x
• C(x) (10200,000) 200,000
• C(x) 40,000,000,000 QR

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The revenue function

• The revenue function which represented in R(x) is
  simply the product of the number of units
  produced x by the price p(x) of the unit.
• So the formula is
• R(x)x . p(x)

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Example

• The demand equation of a certain product is
  p6-1/2x dollars.
• Find the revenue
• R(x) x.p
• x(6-1/2x)
• 6x-1/2x2

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The Profit Function

• The profit function is nothing but the revenue
  function minus the cost function
• So the formula is
• P(x) R(x)-C(x)
• ?x.p(x)-C(x)

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• Notice that
• A maximum profit is reached when
• the first derivative of P(x) when P(x) is zero
  or doesnt exist
• And
• 2. The second derivative of P(x) is always
  negative P?(x)lt0
• when P(x)0 this is the critical point, and
  since P?(x)lt0 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)lt0

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Example

• A factory that produces i-pods computed its
  demand, and costs and came to realize that the
  formulas are as follows
• p100-0.01x
• C(x)50x10000

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• Find
• The number of units that should be produced for
  the factory to obtain maximum profit.
• The price of the unit.

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• R(x)x.p
• x.(100-0.01)
• 100x -0.01x2
• Now
• P(x)R(x)-C(x)
• 100x-0.01x2 (50x10000)
• -0.01x2 50x-10000

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• Now the maximum point in the parabola will occur
  When P'(x)0 P"(x)lt0
• P(x) -0.01x2 50x-10000
• P'(x)-0.02x50
• ? -0.02x500
• ? -0.02x-50
• ? x2500
• P'(2500) -50500
• So the P'(X)0 When x2500
• - Now we will find the second derivative
• P?(x) -0.02
• P?(2500) -0.02lt0
• So x2500 is at a local maximum

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Graph
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• Finally to find the price that is needed to be
  charged per unit we return to the demand
  function
• P100-0.01(2500)
• 100-25 75

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• Thank
• you