Question 15.21: A multiplant monopoly produces the quantities q1, q2 and q3 ......
A multiplant monopoly produces the quantities q_{1} , q_{2} and q_{3} in the three plants that it operates and faces the profit function
π = −24 + 839q_{1} + 837q_{2} + 835q_{3} − 5.05q_{1}^2 − 5.03q_{2}^2 − 5.02q_{3}^2 − 10q_{1}q_{2} − 10q_{1} q_{3} − 10q_{2}q_{3}
Find the output levels in each of its three plants q_{1} , q_{2} and q_{3} that will maximize profit and use the Hessian to check that second-order conditions are met.
Differentiating this π function with respect to q_{1} , q_{2} and q_{3} and setting equal to zero to find the optimum values where the first-order conditions are met, we get:
\pi_{1} = 839 − 10.1q_{1} − 10q_{2} − 10q_{3} = 0 (1)
\pi_{2} = 837 − 10q_{1} − 10.06q_{2} − 10q_{3} = 0 (2)
\pi_{3} = 835 − 10q_{1} − 10q_{2} − 10.04q_{3} = 0 (3)
These conditions can be rearranged to get
839 = 10.1q_{1} + 10q_{2} + 10q_{3}
837 = 10q_{1} + 10.06q_{2} + 10q_{3}
835 = 10q_{1} + 10q_{2} + 10.04q_{3}
These simultaneous equations can be specified in matrix format and solved by the matrix inversion method to get the optimum values of q_{1} , q_{2} and q_{3} as 42, 36.6 and 4.9, respectively. (The full calculations are not set out here as the objective is to explain how the Hessian is used to check second-order conditions, but you can check these answers using Excel if you are not sure how these values are calculated.)
Differentiating (1), (2) and (3) again we can derive the Hessian matrix of second-order partial derivatives
H = \begin{bmatrix} \pi _{11}& \pi _{12}&\pi _{13} \\ \pi _{21}&\pi _{22} &\pi _{23} \\\pi _{31} &\pi _{32} &\pi _{33} \end{bmatrix} = \begin{bmatrix} -10.1& −10& −10 \\ −10& −10.06 &−10 \\ −10& −10& −10.04\end{bmatrix}
The determinants of the three principal minors will therefore be
\left|H_1\right| = \left|\pi_{11}\right| = −10.1
\left|H_2\right| = \begin{vmatrix} \pi _{11}& \pi _{12} \\ \pi _{21}&\pi _{22} \end{vmatrix} = \begin{vmatrix} -10.1& −10 \\ −10& −10.06\end{vmatrix} = 101.606 − 100 = 1.606
\left|H_3\right| = \begin{vmatrix} \pi _{11}& \pi _{12}&\pi _{13} \\ \pi _{21}&\pi _{22} &\pi _{23} \\\pi _{31} &\pi _{32} &\pi _{33} \end{vmatrix} = \begin{vmatrix} -10.1& −10& −10 \\ −10& −10.06 &−10 \\ −10& −10& −10.04\end{vmatrix} = −0.1242
(You can check the H_3 determinant calculations using the Excel MDETERM function.) This Hessian is therefore negative definite as
\left|H_1\right| = −10.1 < 0, \left|H_2\right| = 1.606 > 0, \left|H_3\right| = −0.1242 < 0
and so the second-order conditions for a maximum are met.