Question 11.11: A firm has a budget of £300 to spend on the three inputs x, ......

The budget constraint is

300 − 4x − y − 6z = 0

and the objective function to be maximized is

Q = 24x^{0.3}y^{0.2}z^{0.3}

Thus the Lagrange function is

G = 24x^{0.3}y^{0.2}z^{0.3} + λ(300 − 4x − y − 6z)

Differentiating with respect to each variable and setting equal to zero gives

\frac{\partial G}{\partial x } = 7.2x^{-0.7}y^{0.2}z^{0.3} − 4λ = 0          λ = 1.8x^{-0.7}y^{0.2}z^{0.3}      (1)
\frac{\partial G}{\partial y} = 4.8x^{0.3}y^{-0.8}z^{0.3} − λ = 0          λ = 4.8x^{0.3}y^{-0.8}z^{0.3}        (2)
\frac{\partial G}{\partial z} = 7.2x^{0.3}y^{0.2}z^{-0.7} − 6λ = 0          λ = 1.2x^{0.3}y^{0.2}z^{-0.7}       (3)
\frac{\partial G}{\partial \lambda }  = 300 − 4x − y − 6z = 0            (4)

A simultaneous three-linear-equation system in the three unknowns x, y and z can now be set up if λ is eliminated. There are several ways in which this can be done. In the method used below we set (1) and then (3) equal to (2) to eliminate x and z and then substitute into (4) to solve for y. Whichever method is used, the point of the exercise is to arrive at functions for any two of the unknown variables in terms of the remaining third variable. Thus, setting (1) equal to (2)

1.8x^{-0.7}y^{0.2}z^{0.3} = 4.8x^{0.3}y^{-0.8}z^{0.3}

Multiplying both sides by x^{0.7}y^{0.8} and dividing by z^{0.3} gives

1.8y = 4.8x
0.375y = x         (5)

We have now eliminated z and obtained a function for x in terms of y. Next we need to eliminate x and obtain a function for z in terms of y. To do this we set (2) equal to (3), giving

4.8x^{0.3}y^{-0.8}z^{0.3} = 1.2 x^{0.3}y^{0.2}z^{-0.7}

Multiplying through by z^{0.7}y^{0.8} and dividing by x^{0.3} gives

4.8z = 1.2y
z = 0.25y

Substituting (5) and (6) into the budget constraint (4)

300 − 4(0.375y) − y − 6(0.25y) = 0
                   300 − 1.5y − y − 1.5y = 0
                                                 300 = 4y
                                                    75 = y

Therefore, from (5)

x = 0.375(75) = 28.125

and from (6)

z = 0.25(75) = 18.75

If these optimal values of x, y and z are used then the maximum output will be

Q = 24x^{0.3}y^{0.2}z^{0.3} = 24(28.125)^{0.3}(75)^{0.2}(18.75)^{0.3} = 373.1 units.