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Econ 407: Additional notes on finding critical values of a function.

Maximization of functions with two or more variables subject to a constraint.

Differential calculus is also used to maximize a function subject to constraint. Given a function f(x, y) subject to constraint g(x, y)= k (a constant), a new function F can be formed by (1) setting the constraint equal to zero, (2) multiplying it by (Lagrange multiplier), and (3) adding the product to the original function:

F(x,y) = f(x,y) + [k-g(x, y)]

Here F9x, y, ) is the lagrangian function, f(x, y) is the original or objective function, and g(x, y) is the constraint. Since the constraint is always set equal to zero, the product pk-g(x, y)] also equals zero, and the addition of the term does not change the value of the objective function. Critical values x , y , and , at which the function is optimized, are found by taking the partial derivatives of F with respect to all three independent variables, setting them equal to zero, and solving simultaneously:

F (x, y, ) = 0 F (x, y, ) = 0 F ( x, y, )= 0

second order conditions differ from those of unconstrained optimization.

optimize the function:

z=4x + 3xy + 6y

subject to the constant x + y = 56.

1. Set the constraint equal zero. 56-x-y=0

multiply it by and add it to the objective function to form the lagrangian function Z.

Z = 4x + 3xy + 6y + (56-x-y)

2. Take the first order partials, set them equal to zero, and solve simultaneously.

Z = 8x + 3y- =0

Z = 3x + 12y - = 0

Z = 56 - x-y = 0

Subtracting (5.9) from (5.8)to eliminate gives

5x-9y=0 X=1.8y

substitute x=1.8y in (5.10)

56-1.8-y=0 y = 20

from which we find x = 36 = 348

substitute the critical values

Z=4(36) + 3(36)(20) + 6(20) + (348) (56-20-36)

4(12960)+ 3(720)+ 6(400) +348(0) = 9744

II.

The lagrange multiplier approximates the marginal impact on the objective function caused by a small change in the constant of the constraint. With = 348 in example 9, a one unit increase (decrease) in the constant of the constraint would cause Z to increase(decrease) by 348 units, as is demonstrated in example 10. In a this the lagrange multiplier is akin to a shadow price. In utility maximization subject to a budget constraint, Will estimate the marginal utility of an extra dollar of income.

example 10. To verify that a one unit change in the constant of the constraint will cause a change of 348 units in Z take the original objective function z=4x + 3xy + 6y and optimize it subject to a new constraint x+ y = 57 in which the constant of the constraint is one unit larger.

Z = 4x + 3xy + 6y + (57-x-y)

Z = 8x +3y- = 0

Z= 3x + 12y- = 0

Z= 57-x-y =0

when solved simultaneously this gives

X = 36.64 y = 20.36 =354.2

substituting these values in the lagrangian function give Z = 10 095 which is 351 larger than the old constrained optimum of 9744, close to the approximation of the 348 increment suggested by

III. In section 3.4 the derivative dy/dx was presented as a single symbol denoting the limit of change y / change x as change x approaches zero. The derivative dy/dx may also be treated as a ratio of differentials in which dy is the differential of y and dx is the differential of x. Given a function of a single independent variable y=f(x), the differential of y, dy, measures the change in y resulting from a small change in x, written dx. Given y=2x + 5x + 4, the differential is found by first taking the derivative which is a rate of change.

dy = 4x + 5 (A derivative)

dx

and then mentally multiplying both sides of the equation by dx, which signifies a small change in the independent variable, to obtain dy= (4x + 5) dx is differential

example 11

if y = 4x + 5x -7 then dy/dx =12x + 10x and the differential is dy=(12x + 10x) dx