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Substitution Method

 Restrained Maximisation: Substitution Method

Illustration 9

Infer a manager of a firm is manufacturing two products X and Y seeks to optimise total profit function which is set by the following Equation π = 200X – 4X^2 – XY – 6Y^2 + 190Y. X and Y denotes volume of two products. The manager of the firm countenances the restraints that the total productivity of the two products must parity to 50.

Solution

The condition to be satisfied is X + Y = 50. With this equation we derive the functions of X and Y. That is

                        X         =          50 – Y and
                        Y         =          50 – X

We can consider any of the above to solve the case. Let us take X = 50 –Y and substitute the value of X in the profit function π.

π          =          200(50-Y) – 4(50-Y)^2 – (50-Y)Y – 6Y^2 + 190Y

            =          10000 – 100Y – 4*2500 + Y^2 – 50 + Y^2 – 6Y^2 + 190Y

            =          – 4Y^2 + 90Y

To optimise the above profit function converted into the above non-restrained form we differentiate it with respect to Y and set it parity to zero and solve for Y. Therefore,

                    =          90 – 8Y           =          0
            dY

                                                8Y       =          90

                                                Y         =          90 / 8

                                                Y         =          11.25

By substituting the value of Y = 11.25, we get,

                                    X + Y              =          50

                                    X + 11.25        =          50

                                                X         =          50 – 11.25 = 38.75

Therefore, the given restraint profit will be optimised if the manager of the firm decides to manufacture 38.75 units of the products X and 11.25 units of product Y. We can ascertain the total profits in this restrained maximum condition by substituting the values of X and Y obtained in the given profit function, therefore,

π          =          200X – 4X^2 – XY – 6Y^2 + 190Y

            =          200 * 38.75 – 4(38.75)^2 – 11.25 * 38.75 – 6(11.25)^2 + 190(11.25)

            =          7750 – 4*1502 – 436 – 6*127 + 2138
           
            =          9452 – 6008 – 762

            =          2682

Lagrange Multiplier Technique

Illustration 10

Let us assume the following profit function π = 200X – 4X^2 – XY – 6Y^2 + 190Y as in the above illustration and that the sum of two products X and Y to be 50. By constituting the Lagrange multiplier method, ascertain the value of λ.

Lπ = 200X – 4X^2 – XY – 6Y^2 + 190Y + λ(X+Y – 50)

The derivative of the above X function is as follows.

200 – 8X – Y + λ = 0                          ……Equation (1)

The derivative of the Y function is as follows.

-X – 12Y + 190 + λ = 0                      ……Equation (2)

And the derivative λ function is as follows.

X + Y – 50 = 0                                    …...Equation (3)

To obtain the values of X and Y we have to deduct Equation (2) from (1), we get

10 – 7X + 11Y = 0                              …..Equation (4)

Now multiplying the Equation (3) with 7, we get the

+ 7X + 7Y – 350 = 0                          …..Equation (5) and added to

– 7X + 11Y + 10 = 0                          …..Equation (4)
      18 Y – 340 = 0

18Y = 340;      Y = 340 / 18    =          19

Substituting the value of Y = 19, we get the value of X = 50 – 19 = 31.

And the value of λ can be obtained by equating in the equation (1)

200 – 8(31) -19 + λ = 0

200 – 248 – 19 + λ = 0

λ  = 267 – 200             =          67

Therefore, λ can be understood as marginal profit at the production level of 50 units. It depicts if the firm is needed to produce 49 units instead of 50 units, its profits will drop by 67.

Alternatively, if the firm were to manufacture 51 instead of 50, then the profits would mount by 67.

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