\chapter{Polynomial and rational functions}
\section{Other forms and applications}
\subsection{General form}
If we are given a quadratic function in standard form, we can
expand the square of the binomial:
\begin{example}
Let \(f(x)=3(x-1)^2-7\). Then
\[
\begin{array}{rcll}
f(x) & = & 3(x-1)^2-7\\[1ex]
& = & 3(x^2-2x+1)-7\\[1ex]
& = & 3x^2-6x+3-7\\[1ex]
& = & 3x^2-6x-4
\end{array}.
\]
\end{example}
The equation obtained this way is called the \textit{general form} of the quadratic function.
So the general form is of the type:
| General form of a quadratic function |
|---|
| \(y=ax^2+bx+c\) |
where \(a\neq 0\) and \(b\), \(c\) are any real numbers. In fact, \(a\) is the same number
for both standard and general form (as you can see from the previous example).
But how are \(h\) and \(k\) related to \(a,b,c\)? As we know, \((h,k)\) are the coordinates
of the vertex, and we need to be able to find the vertex if we are given the general form
\(y=ax^2+bx+c\). Below are the formulas to find the vertex.
| Vertex formula |
|---|
| The vertex \(V=(h,k)\)
of the quadratic function \(f(x)=ax^2+bx+c\) is found by |
| \( h=-\dfrac{b}{2a}, \hspace{2ex} k=f(h).\) |
Using the vertex formula, we can easily re-write the function in standard form, and then graph
it as before.
\begin{example}
We find the vertex of the quadratic function \(f(x)=2x^2-4x+1\). We have
\[a=2, \ \ b=-4,\ \ c=1.\]
So
\[
\begin{array}{rcll}
h & = & \displaystyle -\frac{b}{2a}\\[2ex]
\ & = & \displaystyle -\frac{-4}{2(2)}\\[2ex]
\ & = & \displaystyle \frac{4}{4}\\[2ex]
\ & = & 1
\end{array}
\]
and
\[
\begin{array}{rcll}
k & = & f(1)\\[1ex]
\ & = & 2(1)^2-4(1)+1\\[1ex]
\ & = & 2-4+1\\[1ex]
\ & = & -1
\end{array}
\]
So the vertex is \(V(h,k)=(1,-1)\), and the standard form is
\[f(x)=2(x-1)^2-1.\]
\end{example}
Having the standard form of a quadratic function makes it easy to find the \(x\)-intercepts. For example,
in the previous example to solve \(2(x-1)^2-1=0\) we can add 1 to both sides, then divide by 2 and take the square
root. This will give us \(x-1\). Then we just add 1 to get \(x\). But sometimes we may not need the standard
form and we just want to find the
\(x\)-intercepts from the general form. In this case it is easier to
either factor the equation, or use the quadratic formula.
\begin{example}
We find the \(x\)-intercepts of \(f(x)=2x^2-4x-16\).
\[
\begin{array}{rcll}
2x^2-4x-16&=&0 & \mbox{The equation we need to solve}\\[1ex]
x^2-2x-8&=&0 &\mbox{Divide by 2}\\[1ex]
(x+2)(x-4)&=&0& \mbox{Factor}\\[1ex]
x+2=0 \mbox{ or } x-4=0 &&& \mbox{Set each factor to zero}\\[1ex]
x=-2 \mbox{ or } x=4 &&& \mbox{Solve for \(x\)}
\end{array}
\]
So the \(x\)-intercepts are \((-2,0)\) and \((4,0)\).
\end{example}
In the next example we find the \(x\)-intercepts for a quadratic function that does not factor.
\begin{example}
Let \(f(x)=x^2-x-5\). In order to factor this equation we would need two whole numbers that add to \(-1\)
and multiply yo \(-5\), and that is impossible. So we use the quadratic formula, that we display here below.
| The quadratic formula |
|---|
| The solutions of \(ax^2+bx+c=0\) are |
| \( \displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\) |
So for our equation we have
\[a=1, b=-1, c=-5,\]
and we find
\[ x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-5)}}{2(1)}=\frac{1\pm\sqrt{21}}{2}.\]
So the \(x\)-intercepts are \(\left(\dfrac{1-\sqrt{21}}{2},0\right)\),
\(\left(\dfrac{1+\sqrt{21}}{2},0\right)\).
\end{example}
\subsection{Factored form}
It is often useful to write a quadratic function so that the zeros are
immediately clear. For example, in Example 2.1.2, we found that the \(x\)-intercepts of
\(f(x)=2(x+1)^2-8\) are \((-3,0)\) and \((1,0)\). This means that the zeros of \(f(x)\) are
\(-3\) and \(1\), and we can write it in factored form
as:
\[f(x)=2(x+3)(x-1)\]
The general rule for writing a quadratic function in factored form is:
| Factored form of a quadratic function |
|---|
| If the quadratic function \(f(x)=ax^2+bx+c\) has zeros
\(x_1\) and \(x_2\) |
| then the factored form is
\(\displaystyle f(x)=a(x-x_1)(x-x_2)\) |
Note that the coefficient \(a\) must always be included in the factored form. This formula
can be used also when the quadratic function does not factor, and we use the quadratic
formula to find the solutions of \(f(x)=0\).
Of course it could happen that the equation \(f(x)=0\) has no real solutions (this happens
when we find a negative number inside the square root). In this case there is no
factored form for \(f(x)\) using real numbers. Or it could be that there is only one zero, that is, \(x_1=x_2\).
In this case the factored form is \(f(x)=a(x-x_1)^2\).
\subsection{Maximum and minimum}
Remember that the graph of a quadratic function \(f(x)=ax^2+bx+c\) is always a parabola.
A parabola opens up if \(a>0\) and opens down if \(a<0\).
When the parabola opens up,
the vertex is, of course, the lowest point on the graph.
This means that the \(y\)-coordinate of the vertex is the minimum value of the quadratic
function \(f(x)\). It also means that \(f(x)\) has no maximum value, because it continues
to grow indefinitely.
\begin{example}
The parabola in the picture opens up, and the vertex is at \((-1,1)\). So the minimum
value of \(f(x)\) is 1, and the function does not have a maximum value.
\[
\begin{array}{c}
\img{U2_1F1.jpg}{}{10em}{} \\
f(x)=(x+1)^2+1
\end{array}
\]
\end{example}
Similarly, when the parabola opens down, the vertex is
the highest point, so the \(y\)-coordinate is the maximum value of the
quadratic function \(f(x)\). And there is no minimum value, because the graph continues to
go down forever.
\begin{example}
The parabola in the picture opens down, and the vertex is at \((-1,2)\). So the maximum
value of \(f(x)\) is 2, and there is no minimum value.
\[
\begin{array}{c}
\img{U2_1F4.jpg}{}{10em}{}\\
f(x)=-(x+1)^2+2
\end{array}
\]
\end{example}
\subsection{Applications of quadratic functions}
Quadratic functions are often used in application problems from Business, Physics and other
subjects. A typical
example is when we are looking for the maximum or minimum value of some quantity that can
be modeled by a quadratic function.
\begin{example}
The cost (in dollars) to produce and sell a product is given by the formula
\[C(x)=0.12x^2-47x+90,000\]
where \(x\) is the number of items sold. We are interested in finding the minimum cost. The given equation for \(C(x)\)
is a parabola opening up. So the minimum cost will be the \(y\)-coordinate of the vertex. First we find the \(x\)-coordinate.
We find
\[
\begin{array}{rcll}
h & = & \displaystyle -\frac{b}{2a}\\[2ex]
& = & \displaystyle -\frac{-47}{2(0.12)}\\[2ex]
& = & 195.83\\
\end{array}
\]
So the number of items sold that will give the minimum cost is \(h=196\)
(we have rounded the answer to the nearest integer because in this
problem \(x\) must clearly be an integer).
The \(y\)-coordinate of the vertex is
\[
\begin{array}{rcll}
k & = & f(h)\\[2ex]
& = & f(196)\\[2ex]
& = & 0.12(196)^2-47(196)+90\mbox{,}000\\[2ex]
& = & 85\mbox{,}397.92
\end{array}
\]
So the minimum cost is \$\(85\mbox{,}397.92\)
\end{example}
Problems
\problem
Find the standard form of the quadratic functions, then draw an accurate graph.
Label the vertex and all intercepts.
\begin{enumerate}
\item
\(f(x)=-x^2+6x-10\)
\item
\(g(x)=x^2+6x+5\)
\item
\(p(x)=-2x^2-8x-8\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
For \(f(x)=-x^2+6x-10\), we have \(a=-1\), \(b=6\), \(c=-10\). We find the vertex:
\[
\begin{array}{rcll}
h&=&\displaystyle -\frac{b}{2a}\\[2ex]
&=&\displaystyle-\frac{6}{2(-1)}\\[2ex]
&=& 3
\end{array}
\]
and
\[
\begin{array}{rcll}
k&=& f(h)\\[1ex]
&=& f(3)\\[1ex]
&=& -3^2+6(3)-10\\[1ex]
&=& -9+18-10\\[1ex]
&=& -1
\end{array}
\]
So the vertex is \(V=(3,-1)\), and then standard form is
\[f(x)=-(x-3)^2 -1.\]
Now we find the \(y\)-intercept:
\[
\begin{array}{rcll}
f(0)&=& -0^2+6(0)-10\\[1ex]
&=& -10
\end{array}
\]
The \(y\)-intercept is \((0,-10)\).
For the \(x\)-intercepts, we need to solve \(f(x)=0\). It is easier if we use the standard form
instead of the general form.
\[
\begin{array}{rcll}
-(x-3)^2 -1&=& 0 & \mbox{Set \(f(x)=0\)}\\[1ex]
(x-3)^2+1&=& 0 & \mbox{Multiply all by \(-1\)}\\[1ex]
(x-3)^2&=&-1 & \mbox{Subtract \(1\) from both sides}\\[1ex]
x-3 & = & \pm\sqrt{-1} & \mbox{Not a real number}
\end{array}
\]
Since we ran into the square root of a negative number, the quadratic function has no \(x\)-intercepts.
To make a more accurate graph (especially when the function has no \(x\)-intercepts), we find a few extra points:
\[
\begin{array}{rcll}
f(1)&=& -1^2+6(1)-10\\[1ex]
&=& -1+6-10\\[1ex]
&=& -5\\[1ex]
f(2)&=& -2^2+6(2)-10\\[1ex]
&=& -4+12-10\\[1ex]
&=&-2\\[1ex]
f(4)&=& -4^2+6(4)-10\\[1ex]
&=&-2\\[1ex]
f(5)&=& -5^2+6(5)-10\\[1ex]
&=& -25+30-10\\[1ex]
&=& -5
\end{array}
\]
As usual, we gather together all the information we have for the graph:
\[
\begin{array}{c|c|c|c|c}
\mbox{Opens} & \mbox{Vertex} & y\mbox{-intercept} & x\mbox{-intercepts} & \mbox{Extra points}\\
\hline
\mbox{Down} & (3,-1) & (0,-10) & \mbox{None} & (1,-5), (2,-2), (4,-2),(5,-5)
\end{array}
\]
Now we can sketch an accurate graph:
\[
\begin{array}{c}
\img{U2_1F16.jpg}{}{10em}{}\\
f(x)=-x^2+6x-10
\end{array}
\]
\item
For \(g(x)=x^2+6x+5\), we have \(a=1\), \(b=6\), \(c=5\).
Vertex:
\[
\begin{array}{rcll}
h & = & \displaystyle -\frac{b}{2a}\\[2ex]
&=& \displaystyle -\frac{6}{2(1)}\\[2ex]
&=& -3\\
\end{array}
\]
and
\[
\begin{array}{rcll}
k&=& g(h)\\[1ex]
&=& g(-3)\\[1ex]
&=& (-3)^2+6(-3)+5\\[1ex]
&=& 9-18+5\\[1ex]
&=& -4
\end{array}
\]
The vertex is \(V=(-3,-4)\), and the standard form is
\[g(x)=(x+3)^2-4.\]
\(y\)-intercept:
\[
\begin{array}{rcll}
g(0)&=& 0^2+6(0)+5\\[1ex]
&=& 5
\end{array}
\]
The \(y\)-intercept is \((0,5)\).
\(x\)-intercepts: use the standard form.
\[
\begin{array}{rcll}
(x+3)^2-4&=& 0& \mbox{ Set \(g(x)=0\)}\\[1ex]
(x+3)^2&=& 4 & \mbox{ Add \(4\) to both sides}\\[1ex]
x+3 & =& \pm\sqrt{4} & \mbox{ Take the square root}\\[1ex]
x+3 &=& \pm 2 & \mbox{Simplify}\\[1ex]
x+3=2 \mbox{ or } x+3= -2 &&& \\[1ex]
x=-1 \mbox{ or } x=-5 &&& \mbox{Solve for \(x\)}
\end{array}
\]
The \(x\)-intercepts are \((-5,0)\) and \((-1,0)\).
Find a few extra points:
\[
\begin{array}{rcll}
g(-4)&=& (-4)^2+6(-4)+5\\[1ex]
&=& 16-24+5\\[1ex]
&=& -3\\[1ex]
g(-2)&=& (-2)^2+6(-2)+5\\[1ex]
&=& 4-12+5\\[1ex]
&=& -3
\end{array}
\]
\((-4,-3)\) and \((-2,-3)\) are points on the graph.
Summarize the information:
\[
\begin{array}{c|c|c|c|c}
\mbox{Opens} & \mbox{Vertex} & y\mbox{-intercept} & x\mbox{-intercepts} & \mbox{Extra points}\\
\hline
\mbox{Up} & (-3,-4) & (0,5) & (-5,0),(-1,0) & (-4,-3), (-2,-3)
\end{array}
\]
Sketch the graph:
\[
\begin{array}{c}
\img{U2_1F17.jpg}{}{10em}{}\\
f(x)=x^2+6x+5
\end{array}
\]
\item
\(p(x)=-2x^2-8x-8\). So \(a=-2\), \(b=-8\), \(c=-8\).
Vertex:
\[
\begin{array}{rcll}
h &=& \displaystyle-\frac{b}{2a}\\[2ex]
& = & \displaystyle -\frac{-8}{2(-2)}\\[2ex]
& = & \displaystyle \frac{8}{-4}\\[2ex]
& = & -2
\end{array}
\]
and
\[
\begin{array}{rcll}
k& = & p(h)\\[1ex]
& = & p(-2)\\[1ex]
& = & -2(-2)^2-8(-2)-8\\[1ex]
& = & -8+16-8\\[1ex]
& = & 0
\end{array}
\]
The Vertex is \((-2,0)\), and the standard form is
\[p(x)=-2(x+2)^2.\]
\(y\)-intercept:
\[
\begin{array}{rcll}
p(0) & = & -2(0)^2-8(0)-8\\[1ex]
& = & -8\\
\end{array}
\]
The \(y\)-intercept is \((0,-8)\).
\(x\)-intercepts:
We know the vertex \((-2,0)\) is on the \(x\)-axis. So it must be the only \(x\)-intercept.
Extra points:
\[
\begin{array}{rcll}
p(-3)& = & -2(-3)^2-8(-3)-8\\[1ex]
& = & -18+24-8\\[1ex]
& = & -2\\[1ex]
p(-1) & = & -2(-1)^2-8(-1)-8\\[1ex]
& = & -2+8-8\\[1ex]
& = & -2
\end{array}
\]
Extra points are \((-3,-2)\) and \((-1,-2)\).
Summarize the information:
\[
\begin{array}{c|c|c|c|c}
\mbox{Opens} & \mbox{Vertex} & y\mbox{-intercept} & x\mbox{-intercept} & \mbox{Extra points}\\
\hline
\mbox{Down} & (-2,0) & (0,-8) & (-2,0) & (-3,-2), (-1,-2)
\end{array}
\]
Sketch the graph:
\[
\begin{array}{c}
\img{U2_1F18.jpg}{}{10em}{}\\
p(x)=-2x^2-8x-8
\end{array}
\]
\end{enumerate}
\end{sol} \mproblem
Find the standard form of the quadratic functions,
then draw an accurate graph. Label the vertex and all the intercepts.
\begin{enumerate}
\item
\(f(x)=x^2-4x+5\)
\item
\(g(x)=-x^2-2x+3\)
\item
\(p(x)=2x^2-4x+2\)
\end{enumerate}
\problem
Rewrite the quadratic function in factored form:
\begin{enumerate}
\item \(f(x)=3x^2+6x+3\)
\item \(g(x)=2x^2-7x+2\)
\item \(h(x)=-x^2+9x-14\)
\item \(p(x)=3x^2+x+7\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item\(f(x)=3x^2+6x+3\).
Find the zeros:
\[
\begin{array}{rcll}
3x^2+6x+3 & = & 0 & \mbox{set \(f(x)=0\)}\\[1ex]
x^2+2x+1 & = & 0 & \mbox{divide all by 3}\\[1ex]
(x+1)^2 & = & 0 & \mbox{factor}\\[1ex]
x+1 & = & 0 & \mbox{solve for x}\\[1ex]
x & = & -1
\end{array}
\]
We found only one zero, \(x_1=-1\). So the factored form for \(f(x)\) is
\[
\begin{array}{rcll}
f(x) & = & a(x-x_1)^2\\[1ex]
& = & 3(x-(-1))^2\\[1ex]
& = & 3(x+1)^2\\
\end{array}
\]
\item \(g(x)=2x^2-7x+2\). Find the zeros:
\[
\begin{array}{rcll}
2x^2-7x+2 & = & 0 & \mbox{set \(g(x)=0\)}\\[2ex]
x & = &\displaystyle \frac{-(-7)\pm\sqrt{(-7)^2-4(2)(2)}}{2(2)}& \mbox{solve by the quadratic formula}\\[2ex]
x & = & \displaystyle \frac{7\pm\sqrt{49-16}}{4} & \mbox{simplify}\\[2ex]
x & = & \displaystyle \frac{7\pm\sqrt{33}}{4}
\end{array}
\]
We found two zeros: \(x_1=\displaystyle \frac{7+\sqrt{33}}{4}\) and \(\displaystyle x_2=\frac{7-\sqrt{33}}{4}\). So the factored form for \(g(x)\) is
\[
\begin{array}{rcll}
g(x) & = & a(x-x_1)(x-x_2)\\[2ex]
& = & \displaystyle 2\left(x-\frac{7+\sqrt{33}}{4}\right)\left(x-\frac{7-\sqrt{33}}{4}\right)
\end{array}
\]
\item \(h(x)=-x^2+9x-14\).
Find the zeros:
\[
\begin{array}{rcll}
-x^2+9x-14 & = & 0 & \mbox{set \(h(x)=0\)}\\[1ex]
x^2-9x+14 & = & 0 & \mbox{multiply all by \(-1\)}\\[1ex]
(x-2)(x-7) & = & 0 & \mbox{factor}\\[1ex]
x-2=0 & \mbox{or} & x-7 =0 & \mbox{set each factor =0}\\[1ex]
x=2 & \mbox{or} & x=7 & \mbox{solve for x}
\end{array}
\]
We found two zeros: \(x_1=2\) and \(x_2=7\). So the factored form for \(h(x)\) is
\[
\begin{array}{rcll}
h(x) & = & a(x-x_1)(x-x_2)\\[1ex]
& = & -(x-2)(x-7)
\end{array}
\]
\item \(p(x)=3x^2+x+7\).
Find the zeros:
\[
\begin{array}{rcll}
3x^2+x+7 & = & 0 & \mbox{set \(p(x)=0\)}\\[2ex]
x & = &\displaystyle \frac{-1\pm\sqrt{(1)^2-4(7)(3)}}{2(3)}& \mbox{solve by the quadratic formula}\\[2ex]
x & = & \displaystyle \frac{-1\pm\sqrt{-83}}{6} & \mbox{simplify}
\end{array}
\]
There are no real zeros. So the quadratic function does not have a factored form.
\end{enumerate}
\end{sol} \mproblem
Rewrite the quadratic functions in factored form.
\begin{enumerate}
\item \(f(x)=3x^2-9x-30\)
\item \(g(x)=x^2+2x+2\)
\item \(h(x)=-2x^2+12x-18\)
\item \(p(x) =x^2+5x-1 \)
\end{enumerate}
\problem
\begin{enumerate}
\item Find the maximum of the quadratic function\\
\(f(x)=-4x^2-5x+3\)
\item Find the minimum of the quadratic function \(g(x)=2x^2-6x-3\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
We need to find the \(y\)-coordinate of the vertex. We use the vertex formula:
\[V=(h,k)\]
\[h=-\frac{b}{2a}, \ k=f(h)\]
For \(f(x)=-4x^2-5x+3\), we have \(a=-4\), \(b=-5\), and so
\[
\begin{array}{rcll}
h & = & \displaystyle -\frac{b}{2a}\\[2ex]
& = & \displaystyle -\frac{-5}{2(-4)}\\[2ex]
& = & \displaystyle -\frac{5}{8}
\end{array}
\]
and
\[
\begin{array}{rcll}
k & = & f(h)\\[2ex]
& = & \displaystyle f\left(-\frac{5}{8}\right)\\[2ex]
& = & \displaystyle -4\left(-\frac{5}{8}\right)^2-5\left(-\frac{5}{8}\right)+3\\[2ex]
& = & \displaystyle -4\left(\frac{25}{64}\right)+\frac{25}{8}+3\\[2ex]
& = & \displaystyle -\frac{25}{16}+\frac{25}{8}+3\\[2ex]
& = & \displaystyle -\frac{25}{16}+\frac{50}{16}+\frac{48}{16}\\[2ex]
& = & \displaystyle \frac{-25+50+48}{16}\\[2ex]
& = & \displaystyle \frac{73}{16}
\end{array}
\]
So the maximum of the quadratic function is \(\displaystyle{\frac{73}{16}}\).
\item
For \(g(x)=2x^2-6x-3\), we have \(a=2\), \(b=-6\), and so
\[
\begin{array}{rcll}
h & = & \displaystyle -\frac{b}{2a}\\[2ex]
& = & \displaystyle -\frac{-6}{2(2)}\\[2ex]
& = & \displaystyle \frac{6}{4}\\[2ex]
& = & \displaystyle \frac{3}{2}
\end{array}
\]
and
\[
\begin{array}{rcll}
k & = & g(h)\\[2ex]
& = & \displaystyle g\left(\frac{3}{2}\right)\\[2ex]
& = & \displaystyle 2\left(\frac{3}{2}\right)^2-6\left(\frac{3}{2}\right)-3\\[2ex]
& = & \displaystyle 2\left(\frac{9}{4}\right)-9-3\\[2ex]
& = & \displaystyle \frac{9}{2}-12\\[2ex]
& = & \displaystyle \frac{9}{2}-\frac{24}{2}\\[2ex]
& = & \displaystyle \frac{9-24}{2}\\[2ex]
& = & \displaystyle -\frac{15}{2}
\end{array}
\]
So the minimum of the quadratic function is \(\displaystyle{-\frac{15}{2}}\)
\end{enumerate}
\end{sol} \mproblem
\begin{enumerate}
\item
Find the minimum of the quadratic function \(f(x)=6x^2-12x+18\)
\item
Find the maximum of the quadratic function \(g(x)=-3x^2+5x+1\)
\end{enumerate}
\problem
The profit (in dollars) from the sale of a product is given by the formula
\[P(x)=-0.15x^2+700x-35\mbox{,}000\]
where \(x\) is the number of items sold.
\begin{enumerate}
\item
Find the number of items sold that will give the maximum profit.
\item
Find the maximum profit.
\end{enumerate}
\begin{sol}
The given formula for the profit \(P(x)\) is a quadratic function with \(a=-0.15<0\).
So the graph is a parabola opening down and the vertex \(V=(h,k)\) is the highest point.
The maximum value of the function \(P(x)\) (that is, the maximum profit) is the \(y\)-coordinate
of the vertex, and the corresponding number of items sold to get the maximum profit
is the \(x\)-coordinate of the vertex.
\begin{enumerate}
\item
The \(x\)-coordinate of the vertex is
\[
\begin{array}{rcll}
h & = & \displaystyle -\frac{b}{2a}\\[2ex]
& = & \displaystyle -\frac{700}{2(-0.15)}\\[2ex]
& = & 2333.33
\end{array}
\]
So the number of items sold that will give the maximum profit is \(2333\).
\item
The \(y\)-coordinate of the vertex is
\[
\begin{array}{rcll}
k & = & f(h)\\[1ex]
& = & f(2333)\\[1ex]
& = & -0.15(2333)^2+700(2333)-35\mbox{,}000\\[1ex]
& = & 781\mbox{,}667
\end{array}
\]
So the maximum profit is \$\(781\mbox{,}667\)
\end{enumerate}
\end{sol} \mproblem
The cost (in dollars) to produce and sell a new model of cell phone is given by the function
\[C(x)=2.5x^2-80x+25\mbox{,}000\]
where \(x\) is the number of cell phones produced and sold.
\begin{enumerate}
\item
Find the number of cell phones that will give the minimum cost.
\item
Find the minimum cost.
\end{enumerate}