Maximisation or minimisation of an objective function when there are no constraints.

However, consumers and managers of business firms quite often face decision problems when there are constraints which limit the choice available to them for optimisation.

For example, a business firm may face a constraint with regard to the limited availability of some crucial raw material, skilled manpower.

Constraint may be of the limited productive capacity as determined by the size of the plant and capital equipment installed. Besides, quite often marketing managers are required to maximise sales subject to the constraint of a given advertising expenditure at their disposal. Further, financial managers also work under the constraint of a given investment requirements of a firm and they are required to minimise the cost of raising capital for that purpose. Managers of business firms also face legal and environmental constraints.

In all these cases when individuals face constraints in their decision making to maximise or minimise their objective functions we have a problem of constrained optimisation. It is worth nothing that the existence of constraints prevents the achievement of the unconstrained optimal. There are two techniques of solving the constrained optimisation problem. (1) Substitution method, (2) Lagrangian multiplier technique. We explain them below.

#### Constrained Optimasation: Substitution Method:

Substitution method to solve constrained optimisation problem is used when constraint equation is simple and not too complex. For example substitution method to maximise or minimise the objective function is used when it is subject to only one constraint equation of a very simple nature.

In this method, we solve the constraint equation for one of the decision variables and substitute that variable in the objective function that is to be maximised or minimised. In this way this method converts the constrained optimisation problem into one of unconstrained optimisation problems of maximisation or minimisation.

An example will clarify the use of substitution method to solve constrained optimisation problem. Suppose a manager of a firm which is producing two products x and y, seeks to maximise total profits function which is given by the following equation

π = 50x – 2x2 – xy – 3y2 + 95y

Where x and y represent the quantities of the two products. The manager of the firm faces the constraints that the total output of the two products must be equal to 25. That is, according to the constraint,

x + y = 25

To solve this constrained optimisation problem through substitution we first solve the constraint equation for x. Thus

x = 25 – y

The next step in the substitution method is to substitute this value of x = 25 – y in the objective function (i.e. the given profit function) which has to be maximised.

Thus, substituting the constrained value of x in the profit function we have:

π = 50 (25 – y) – 2 (25 – y) 2 – (25 – y) y – 3y2 + 95y

= 1250 – 50y – 2 (625 / 50y + y2) – 25y + y2 – 3y2 + 95y

= 1250 – 50y – 1250 + 100y – 2y2 – 25y + y2 – 3y2 + 95y

= 12y – 4y2

To maximise the above profit function converted into the above unconstrained form we differentiate it with respect to y and set it equal to zero and solve for y. Thus,

dπ / dy =120 – 8y = 0

or, 8y = 120

y = 15

Substituting y = 15 in the given constraint equation (x + y = 25) we have

X + 15 =25

X = 25- 15 =10

Thus, given the constraint, profit will be maximised if the manager of the firm decides to produce 10 units of the product x and 15 units of the product y. We can find the total profits in this constrained optimum situation by substituting the values of x and y obtained in the given profit function. Thus,

π = 50x- 2x2 – xy – 3y2 + 95y

π = 50 x 10 – 2(10)2 – 10 x 15 – 3(15)2 + 95 x 15

= 500 – 200 – 150 – 675 + 1425

= 1925 – 1025 = 900

#### Lagrange Multiplier Technique:

The substitution method for solving constrained optimisation problem cannot be used easily when the constraint equation is very complex and therefore cannot be solved for one of the decision variable. In such cases of constrained optimisation we employ the Lagrangian Multiplier technique. In this Lagrangian technique of solving constrained optimisation problem, a combined equation called Lagrangian function is formed which incorporates both the original objective function and constraint equation.

This Lagrangian function is formed in a way which ensures that when it is maximised or minimised, the original given objective function is also maximised or minimised and at the same time it fulfills all the constraint requirements. In creating this Lagrangian function, an artificial variable λ (Greek letter Lamda) is used and it is multiplied by the given constraint function having been set equal to zero. λ is known as Lagrangian multiplier.

Lagrangian Multiplier:

Since Lagrangian function incorporates the constraint equation into the objective function, it can be considered as unconstrained optimisation problem and solved accordingly. Let us illustrate Lagrangian multiplier technique by taking the constrained optimisation problem solved above by substitution method.

In that problem manager of a firm was to maximise the following profit function:

π = 50x – 2x2– xy – 3y2 + 95y subject to the constraint,

x + y = 25

Where x and y are the outputs of two products produced by the firm.

In order to constitute Lagrangian function we first set the constraint function equal to zero by bringing all the terms to the left side of the equation. In doing so we have

x + y – 25 = 0

The next step in creating the Lagrangian function is to multiply this form of the constraint function by the unknown artificial factor λ and then adding the result to the given original objective function. Thus combining the constraint and the objective function through Lagrangian multplier (λ) we have

Lπ = 50x – 2x2 – xy – 3y2 + 95y + λ (x + y – 25)

Where Lπ stands for the expression of Lagrangian function.

As stated above, the Lagrangaion function can be considered as unconstrained optimisation function. It will be seen that in this function there are three unknowns x, y and λ. Note that the solution that maximises Lagrangian function (Lπ) will also maximise profit (λ) function:

For maximising Lπ, we first find partial derivatives of Lπ with respect to three unknown x, y and λ and then set them equal to zero. Thus

∂Lπ/δ = 50 – 4x – y + λ =0 …..(i)

∂Lπ /δy = – x – 6y + 95 + λ =0 …..(ii)

∂Lπ /δλ = x + y – 25 = 0 ….(iii)

Note that the last equation (iii) is the constraint subject to which the original profit function has to be maximised. In fact, Lagrangian function is so constructed that the partial derivative of Lπ with respect to λ (Lagrangian multiplier) always yields the original constraint function. Further, since the partial derivative of Lπ with respect to λ is set equal to zero, it not only ensures that the constraint of the optimisation problem is fulfilled but also converts the Lagrangian function into original constrained profit maximisation problem so that the solution to both of them will yield the same result.

Let us now solve the above system of three equations with three unknowns to find the optimal values of x and y. To do so we first substract the equation (ii) from equation (i) and get

– 45 – 3x + 5y = 0 ….(iv)

Now, multiplying equation (iii) by 3 and adding it to equation (iv) we have

– 45 – 3x + 5y =0

-75 + 3x + 3y =0/ -120 + 8y – =0

Thus, 8y = 120

y= 15

Substituting the value of y = 15 in the constraint function x + y = 25 we get x equal to 10. Thus, we find the same optimal values of x and y as we obtained in the substitution method. Lagrange technique of solving constrained optimisation is highly significant for two reasons.

First, as noted above, when constraint conditions are too many or too complex, it is not feasible to use substitution method’ and therefore in such cases it is easy to use Lagrange technique for solution of constrained optimisation problems.

Second, the use of Lagrangian multiplier technique also provides the decision maker as additional important information.

This information is the value of λ that is, Lagrangian multiplier itself. The value of λ has a significant economic interpretation. The value of λ shows the marginal effect on the solution of the objective function when there is a unit change in the constraint. Thus, in our example of profit maximization, value of λ indicates the marginal profit caused by a one unit change in the output of products, that is, change in the total profit when output constraint either increases from 25 to 26 or decreases from 25 to 24.

The value of λ can be obtained by substituting the solved values of x and y in a partial derivative equation containing λ in the Lagrangian function. Thus, in our above example, the value of X can be obtained by substituting x = 10 and y = 15 in equation (ii) above

Thus, in the above equation (ii)

– x – 6y + 95 + λ =0

λ = x + 6y – 95

λ = 10 + 90 – 95

λ = +5

Here λ can be interpreted as marginal profit at the production level of 25 units. It shows if the firm is required to produce 24 units instead of 25 units, its profits will fall by 5. On the other hand, if firm were to produce 26 instead of 25 units, its profits will increase by about 5.

Note that Lagrange technique maximises profits under a constraint. It does not solve unconstrained profit maximization.