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# System prepended metadata

title: PID-334

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For the first question:
a) We can form the following matrix equations:
$$\begin{bmatrix} 1 & 1 & 1 \\ 2 & 6 & 6 \\ 12 & 20 & 8 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1000 \\ 4400 \\ 13200 \end{bmatrix},$$

where $x$ is the number of acres of land, $y$ is the number of hours of labor, and $z$ is the cost of seed.

b) The matrix form of the equations in (a) is $$AZ=B,$$where $A$ is the coefficient matrix, $Z$ is the matrix of variables, and $B$ is the matrix of constants:
$$\begin{bmatrix} 1 & 1 & 1 \\ 2 & 6 & 6 \\ 12 & 20 & 8 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1000 \\ 4400 \\ 13200 \end{bmatrix}$$

c) To solve the equations using Cramer's rule, we need to compute the determinant of the coefficient matrix and the determinants of the matrices obtained by replacing each column of the coefficient matrix with the matrix of constants. Then, we can solve for the variables using the following formulas:
$$x = \frac{\begin{vmatrix} b_1 & 1 & 1 \\ b_2 & 6 & 6 \\ b_3 & 20 & 8 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 1 \\ 2 & 6 & 6 \\ 12 & 20 & 8 \end{vmatrix}}=\frac{\begin{vmatrix} 1000 & 1 & 1 \\ 4400 & 6 & 6 \\ 13200 & 20 & 8 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 1 \\ 2 & 6 & 6 \\ 12 & 20 & 8 \end{vmatrix}}=\frac{
-19200}{-48}=
400$$

$$y = \frac{\begin{vmatrix} 1 & b_1 & 1 \\ 2 & b_2 & 6 \\ 12 & b_3 & 8 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 1 \\ 2 & 6 & 6 \\ 12 & 20 & 8 \end{vmatrix}}=\frac{\begin{vmatrix} 1 & 1000 & 1 \\ 2 & 4400 & 6 \\ 12 & 13200 & 8 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 1 \\ 2 & 6 & 6 \\ 12 & 20 & 8 \end{vmatrix}}=\frac{-14400
}{-48}=300$$

$$z = \frac{\begin{vmatrix} 1 & 1 & b_1 \\ 2 & 6 & b_2 \\ 12 & 20 & b_3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 1 \\ 2 & 6 & 6 \\ 12 & 20 & 8 \end{vmatrix}}=\frac{\begin{vmatrix} 1 & 1 & 1000 \\ 2 & 6 & 4400 \\ 12 & 20 & 13200 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 1 \\ 2 & 6 & 6 \\ 12 & 20 & 8 \end{vmatrix}}=\frac{-14400}{-48}=300$$

Thus, $x=400,$ $y=300$ and $z=300$.

**For the second question:**

a) To find the market equilibrium price and quantity, we need to set the demand function equal to the supply function and solve for $Q$:
$$1000 - 25Q = 100 + Q^2$$

Rearranging and simplifying, we get:
$$Q^2 + 25Q - 900 = 0$$

Solving for $Q$ using the quadratic formula, we get:
$$Q = \frac{-25 \pm \sqrt{25^2 + 4 \cdot 900}}{2}$$

$$Q = 20$$  $$Q=-45$$

Since the quantity cannot be negative, the market equilibrium quantity is $Q = 20$.

To find the market equilibrium price, we can substitute $Q = 20$ into either the demand or supply function:
$$P = 1000 - 25Q = 1000 - 25 \cdot 20 = 500$$

Therefore, the market equilibrium price is $P = 500,$ and the market equilibrium quantity is $Q = 20$.

b) The consumer surplus is the difference between the maximum price that consumers are willing to pay and the market price, integrated over the range of quantities demanded. In this case, the maximum price that consumers are willing to pay is given by the demand function, which is $$P = 1000 - 25Q.$$ At the market equilibrium quantity of $Q = 20,$ the market price is $P = 500$. Therefore, the consumer surplus is:
$$\int_0^{20} (1000 - 25Q - 500) dQ = \int_0^{20} (500 - 25Q) dQ = \left[500Q - \frac{25}{2}Q^2\right]_0^{20} = 5000$$

The producer surplus is the difference between the market price and the minimum price that producers are willing to accept, integrated over the range of quantities supplied. In this case, the minimum price that producers are willing to accept is given by the supply function, which is $$P = 100 + Q^2$$. At the market equilibrium quantity of $Q = 20$, the market price is $P = 500$. Therefore, the producer surplus is:
$$\int_0^{20} (500 - 100 - Q^2) dQ = \int_0^{20} (400 - Q^2) dQ = \left[400Q - \frac{1}{3}Q^3\right]_0^{20} = 5333.3.$$