Question 15.20: A firm has the production function TC = 120 + 0.1q² and sell......

From the two demand schedules we can derive

p_{1} = 400 − 0.5q_{1}        TR_{1} = 400q_{1} − 0.5q_{1}^2       MR_{1} = 400 − q_{1}
p_{2} = 300 − 0.4q_{2}        TR_{2} = 300q_{2} − 0.4q_{2}^2      MR_{2} = 300 − 0.8q_{2}

Given that total output q = q_{1} + q_{2} then

TC = 120 + 0.1q² = 120 + 0.1(q_{1} + q_{2} )²
      = 120 + 0.1q_{1}^2 + 0.2q_{1}q_{2} + 0.1q_{2}^2

Therefore

π = TR_{1} + TR_{2} − TC
   = 400q_{1} − 0.5q_{1} ^2 + 300q_{2} − 0.4q_{2}^2 − 120 − 0.1q_{1}^2 − 0.2q_{1}q_{2} − 0.1q_{2}^2
    = 400q_{1} − 0.6q_{1}^2 + 300q_{2} − 0.5q_{2}^2 − 120 − 0.2q_{1}q_{2}

FOC for a maximum require

\frac{\partial \pi }{\partial q_1} = 400 − 1.2q_{1} − 0.2q_{2} = 0      therefore      400 = 1.2q_{1} + 0.2q_{2}         (1)
\frac{\partial \pi }{\partial q_2} = 300 − q_{2} − 0.2q_{1} = 0       therefore     300 = 0.2q_{1} + q_{2}        (2)

To find the optimum values that satisfy the FOC, the simultaneous equations (1) and (2) can be set up in matrix format as

Aq = \begin{bmatrix}1.2 &0.2 \\ 0.2 &1 \end{bmatrix}\begin{bmatrix}q_1 \\ q_2 \end{bmatrix} =  \begin{bmatrix}400\\ 300 \end{bmatrix} = b

Using Cramer’s rule to solve for the sales in each market gives

q_1  =  \frac{\begin{vmatrix} 400 &0.2\\ 300 &1 \end{vmatrix} }{\begin{vmatrix}1.2 &0.2 \\ 0.2 &1 \end{vmatrix} }  =  \frac{400  –   60}{1.2  –  0.04}  =  \frac{340}{1.16} = 293.1
q_2  =  \frac{\begin{vmatrix} 1.2 &400\\ 0.2 &300 \end{vmatrix} }{\begin{vmatrix}1.2 &0.2 \\ 0.2 &1 \end{vmatrix} } = \frac{360  –  80}{1.2  –  0.04}  =  \frac{280}{1.16} = 241.4

To check the second-order conditions we return to the first-order partial derivatives and then find the second-order partial derivatives and the cross partial derivatives. Thus, from

\frac{\partial \pi }{\partial q_1} = 400 − 1.2q_{1} − 0.2q_{2}    and    \frac{\partial \pi }{\partial q_2} = 300 − q_{2} − 0.2q_{1}

we get

\frac{\partial^2 \pi }{\partial q_1^2} = −1.2      \frac{\partial^2 \pi }{\partial q_1\partial q_2} = −0.2       \frac{\partial ^2\pi }{\partial q_2^2} = −1       \frac{\partial^2 \pi }{\partial q_2\partial q_1} = −0.2

The Hessian matrix is therefore

H = \begin{bmatrix} \pi_{11} & \pi_{12} \\ \pi_{21} & \pi_{22} \end{bmatrix}   =  \begin{bmatrix}-1.2 & -0.2 \\ -0.2 & -1 \end{bmatrix}

and the determinants of the principal minors are

\left|H_1\right| = −1.2 < 0

and

\left|H_2\right| = \begin{vmatrix}-1.2 &-0.2 \\ -0.2 & -1 \end{vmatrix}   = 1.2 − 0.04 = 1.16 > 0

As \left|H_1\right| < 0 and \left|H_2\right| > 0 the Hessian is negative definite. Therefore SOC for a maximum are met.