Quiz 5 (Solution)

Date: MATH 1090-4 - Fall 2003

1. Suppose that the profit from a sale of $x$ units of a product is $P(x) = 6400x-18x^2-400$.
   1. What level(s) of production yield a profit of $ 61800?

      We have to solve $61800=P(x)$ for $x$. That is, we need to solve $61800= 6400x-18x^2-400$, or, $18x^2-6400x+62200=0$. Using the quadratic formula we find

      $$\begin{split}x &= \frac{6400 \pm \sqrt{(-6400)^2 - 4\cdot 18\... ...m 6040}{36} = \begin{cases}345.56\\ 10. \end{cases} \end{split}$$

      
      

      Thus, the given profit is achieved at two levels of production: $x=10$ and $x=345.56$.

   2. Can a profit of more than $ 61800 be made?

      Yes. We can see this in two ways.

      First way. A profit can be made if we can solve the quadratic equation for the given profit. If, in the above equation we increase $61800$ to something higher, say $62000$, the number inside the square root is still positive and so the equation can be solved.

      Second way. We see that the profit function is a quadratic function with negative coefficient, therefore, the vertex of the parabola is a maximum (the maximum profit). We find the vertex using the formula $(-\frac{b}{2a}, P(-\frac{b}{2a})) = (177.78,568488.89)$, so that the maximum profit is $568488.89$. Thus, any profit between $61800$ and $568488.89$ can be made.

2. For the graph of $f(x)=4x^2+4x-15$:
   1. Find the coordinates of the vertex.

      The vertex has coordinates

      $$(-\frac{b}{2a},f(-\frac{b}{2a})) = (-\frac{4}{8}, f(-\frac{1}{2})) = (-\frac{1}{2},-16).$$

      
      

   2. Is the vertex a maximum or minimum of $f$?

      Since the quadratic function has positive coefficient of $x^2$, the parabola points upward and the vertex corresponds to a minimum.

   3. Find all the intercepts.

      The $y$-intercept comes from evaluating $f(0) = -15$, so it is the point $(0,-15)$.

      To find the $x$-intercepts we have to solve $f(x)=0$, that is, $4x^2+4x-15 = 0$. Using the quadratic formula:

      $$x = \frac{-4\pm \sqrt{4^2-4\cdot 4\cdot (-15)}}{2\cdot 8} = \frac{-4\pm 16}{8} = \begin{cases}\frac{3}{2}\\ -\frac{5}{2}. \end{cases}$$

      
      
      Therefore, the $x$-intercepts are the points $(\frac{3}{2},0)$ and $(-\frac{5}{2}, 0)$.

   4. Using all the information that you collected so far, sketch the graph of $f$.

      The graph is shown in Figure 1.

      Figure 1: Graph of $f(x)=4x^2+4x-15$