Quiz 6 (Solution)

Date: MATH 1090-4 - Fall 2003

1. If the supply function for a commodity is $p=q^2+8q+16$ and the demand function is $p=-3q^2+6q+436$, find the equilibrium quantity and equilibrium price.

   The equilibrium quantity and price are the (positive) solution(s) of the system

   $$\begin{cases}p=q^2+8q+16\\ p=-3q^2+6q+436. \end{cases}$$

   
   

   We solve $q^2+8q+16 = -3q^2+6q+436$, so that $4q^2+2q-420 = 0$. Then, using the quadratic formula,

   $$q = \frac{-2\pm \sqrt{4-4\cdot4\cdot(-420)}}{8} = \frac{-2\pm\sqrt{6724}}{8} = \frac{-2\pm 82}{8} = \begin{cases}10\\ -10.5 \end{cases}$$

   
   

   Since negative quantities are not possible, the equilibrium quantity is $q=10$.

   Then, using the supply function we find $p = q^2+8q+16 = 10^2+8\cdot10+16 = 196$, so that the equilibrium price is $p=196$.

2. A company has fixed costs of $ 28000 and variable costs of $0.4x+222$ dollars per unit, where $x$ is the total number of units produced. Also, the selling price of its product is $1250 - 0.6x$ dollars per unit.
   1. Write the total cost function $C(x)$.

      We have: total cost $=$ (fixed cost) $+$ (variable cost) $=$ (fixed cost) $+$ (unit variable cost) $\times$ (units produced). In other words:

      $$C(x) = 28000 + (0.4x+222) x = 28000+222x+0.4x^2.$$

      
      

   2. Write the total revenue function $R(x)$.

      We have: total revenue $=$ (unit price) $\times$ (units produced). In other words:

      $$R(x) = (1250-0.6x)x = 1250x -0.6x^2.$$

      
      

   3. Write the total profit function $P(x)$.

      $$P(x) = R(x)-C(x) = 1250x -0.6x^2 - (28000+222x+0.4x^2) = -28000 + 1028x -x^2.$$

      
      

   4. Find the break even point(s).

      We have to solve $C(x)=R(x)$ (or, equivalently, $P(x)=0$). We have $1250x -0.6x^2 = 28000+222x+0.4x^2$ so that $28000 - 1028x +x^2 = 0$. Then, using the quadratic formula we obtain

      $$x = \frac{1028 \pm \sqrt{(-1028)^2-4\cdot 28000}}{2} = \frac{1028 \pm \sqrt{944784}}{2} = \frac{1028\pm 972}{2} \begin{cases}1000\\ 28. \end{cases}$$

      
      
      The break even points happen when the production is $x=1000$ or $x=28$.

   5. Find the maximum revenue.

      The revenue $R(x)$ is represented by a parabola pointing down, so that the vertex of the parabola computes the maximum revenue.

      We find the vertex $(-\frac{b}{2a}, R(-\frac{b}{2a})) = (-\frac{1250}{-1.2}, R(-\frac{1250}{-1.2})) = (1041.66, 651041.66)$.

      The maximum revenue is 651041.66.