The local theater group sold tickets to its opening-night performance for $5 and drew an audience of 100 people. The next night they reduced the ticket price by $0.25 and 10 more people attended; that is, 110 people bought tickets at $4.75 apiece. In fact, for each $0.25 reduction in ticket price, 10 additional tickets can be sold.
Enter your expressions for the price of a ticket, the number of tickets sold, and the total revenue into the calculator as \(Y_1, ~Y_2,\) and \(Y_3\text{.}\) Use the Table feature to verify that your algebraic expressions agree with your table from part (1).
Use your calculator to graph your expression for total revenue in terms of \(x\text{.}\) Use your table to choose appropriate window settings that show the high point of the graph and both \(x\)-intercepts.
What is the maximum revenue possible from ticket sales? What price should the theater group charge for a ticket to generate that revenue? How many tickets will the group sell at that price?
The vertex of a parabola \(~y=ax^2+bx~\) is an interesting point. It is the highest or lowest point on the parabola, and it "anchors" the parabola’s position in the plane. As we saw in the Investigation, the vertex may represent the maximum or minimum possible value for a variable, so it has important implications for applications. Thus, it will be very useful to have a way to find the vertex easily.
So the coordinates of the vertex are \(\left(\dfrac{1}{4},\dfrac{9}{8}\right)\text{.}\) Alternatively, we can use the calculator to evaluate \(-2x^2 + x + 1\) for \(x = 0.25\text{.}\) We enter
and press ENTER. The calculator returns the \(y\)-value \(1.125\text{.}\) Thus, the vertex is the point \((0.25, 1.125)\text{,}\) which is the decimal equivalent of \(\left(\dfrac{1}{4},\dfrac{9}{8}\right)\text{.}\)
We set \(x=0\) to find the \(y\)-intercept, \((0,1)\text{.}\) Now, the axis of symmetry of the parabola is the vertical line that passes through its vertex. That line is \(x=-1.5\text{,}\) so the \(y\)-intercept lies 1.5 units to the right of the axis. There must be another point on the parabola with the same \(y\)-coordinate as the intercept but 1.5 units to the left of the axis. This point is \((-3,1)\text{.}\)
Many quadratic models arise as the product of two variables, one of which increases while the other decreases. For example, in the Investigation for Section 3.3 we looked at the areas of different rectangles with the same perimeter. The area of a rectangle is the product of its length and its width, or \(A=lw\text{.}\) If we require that the rectangle have a certain perimeter, then as we increase its length, we must also decrease its width.
Finding the maximum or minimum value for a variable expression is a common problem in applications. For example, if you own a company that manufactures blue jeans, you might like to know how much to charge for your jeans in order to maximize your revenue. As you increase the price of the jeans, your revenue may increase for a while. But if you charge too much for the jeans, consumers will not buy as many pairs, and your revenue may actually start to decrease. Is there some optimum price you should charge for a pair of jeans in order to achieve the greatest revenue?
Late Nite Blues finds that it can sell \(600 - 15x\) pairs of jeans per week if it charges \(x\) dollars per pair. (Notice that as the price increases, the number of pairs of jeans sold decreases.)
Write an equation for the revenue as a function of the price of a pair of jeans.
Using the formula for revenue stated above, we find
\begin{align*}
\text{Revenue} \amp= (\text{price of one item})(\text{number of items sold})\\
R \amp = x(600 - 15x)\\
R \amp = 600x - 15x^2
\end{align*}
The revenue takes on its maximum value when \(x = 20\text{,}\) and the maximum value is \(R = 6000\text{.}\) This means that Late Nite Blues should charge $20 for a pair of jeans in order to maximize revenue at $6000 a week.
If the equation relating two variables is quadratic, then the maximum or minimum value is easy to find: It is the value at the vertex. If the parabola opens downward, as in Example 4.2.5, there is a maximum value at the vertex. If the parabola opens upward, there is a minimum value at the vertex.
Now notice that the coordinates of the vertex, \((3,-8)\text{,}\) are apparent in the original equation; we don’t need to do any computation to find the vertex.
We compare the equation to the vertex form to see that the coordinates of the vertex are \((4, 6)\text{.}\) For this equation, \(a = -3 \lt 0\text{,}\) so the parabola opens downward. The vertex is the maximum point of the graph.
which confirms that when \(x = 4\text{,}\)\(y = 6\text{.}\) Next, notice that if \(x\) is any number except \(4\text{,}\) the expression \(-3(x - 4)^2\) is negative, so \(y \lt 6\text{.}\) Therefore, \(6\) is the maximum value for \(y\) on the graph, so \((4, 6)\) is the high point or vertex.
You can also rewrite \(y = -3(x - 4)^2 + 6\) in standard form and use the formula \(x_v = \dfrac{-b}{2a}\) to confirm that the vertex is the point \((4, 6)\text{.}\)
Any quadratic equation in vertex form can be written in standard form by expanding, and any quadratic equation in standard form can be put into vertex form by completing the square. \(~\alert{\text{[TK]}}\)
We must add \(1\) to complete the square. However, we are really adding \(3(1)\) to the right side of the equation, so we must also subtract \(3\) to compensate:
\begin{equation*}
y = 3(x^2 - 2x \alert{{}+{}1}) - 1\,\alert{-\,3}
\end{equation*}
The expression inside parentheses is now a perfect square, and the vertex form is
\begin{equation*}
y = 3(x - 1)^2 - 4
\end{equation*}
The vertex is the point \((-3,2)\text{.}\) We can find the \(y\)-intercept by setting \(x=0\text{.}\)
\begin{equation*}
y = \dfrac{-1}{2}(\alert{0}+3)^2-2 = \dfrac{-9}{2}-2 = -6\dfrac{1}{2}
\end{equation*}
The \(y\)-intercept is the point \(\left(0, -6\dfrac{1}{2}\right)\text{.}\) The axis of symmetry is the vertical line \(x=-3\text{,}\) and there is a symmetric point equidistant from the axis, namely \(\left(-6, -6\dfrac{1}{2}\right)\text{.}\) We plot these three points and sketch the parabola through them. \(~\alert{\text{[TK]}}\)
When Andre practices free-throws at the park, the ball leaves his hands at a height of 7 feet, and reaches the vertex of its trajectory 10 feet away at a height of 11 feet.
Find a quadratic equation for the ball’s trajectory.
If Andre’s feet are at the origin, then the vertex of the ball’s trajectory is the point \((\alert{10},\alert{11})\text{,}\) and its \(y\)-intercept is \((0,7)\text{.}\) We start with the vertex form for a parabola.
\begin{align*}
y \amp =a(x-x_v)^2+y_v\\
y \amp =a(x-\alert{10})^2+\alert{11}
\end{align*}
We use the point \((0,7)\) to find the value of \(a\text{.}\)
We’d like to know if the point \((15,10)\) is on the trajectory of Andre’s free-throw. We substitute \(x=\alert{15}\) into the equation.
\begin{align*}
y \amp =-0.04(\alert{15}-10)^2+11\\
y \amp =-0.04(25)+11 = 10
\end{align*}
From our computations, we see that the point \((15,10)\) is indeed on the trajectory. However, because Andre’s shot will probably hit the backboard just where the hoop attaches and bounce off, so it is unlikely that his shot will score.
The monthly dues for Rafael’s condo association are $120. However, for each new tenant who moves in, the dues will be reduced by $5. Write an expression for the dues if \(x\) new tenants move in.
Gavin has rented space for a booth at the county fair. As part of his display, he wants to rope off a rectangular area with 80 yards of rope.
Let \(w\) represent the width of the roped-off rectangle, and write an expression for its length. Then write an expression in terms of \(w\) for the area \(A\) of the roped-off space.
The owner of a motel has 60 rooms to rent. She finds that if she charges $20 per room per night, all the rooms will be rented. For every $2 that she increases the price of a room, three rooms will stand vacant.
Complete the table. The first two rows are filled in for you.
Let stand for the number of $2 price increases the owner makes. Write algebraic expressions for the price of a room, the number of rooms that will be rented, and the total revenue earned at that price.
Use your calculator to make a table of values for your algebraic expressions. Let \(Y_1\) stand for the price of a room, \(Y_2\) for the number of rooms rented, and \(Y_3\) for the total revenue. Verify the values you calculated in part (a).
What is the lowest price that the owner can charge for a room if she wants her revenue to exceed $1296 per night? What is the highest price she can charge to obtain this revenue?
What is the maximum revenue the owner can earn in one night? How much should she charge for a room to maximize her revenue? How many rooms will she rent at that price?
A travel agent offers a group rate of $2400 per person for a week in London if 16 people sign up for the tour. For each additional person who signs up, the price per person is reduced by $100.
Let \(x\) represent the number of additional people who sign up. Write expressions for the number of people signed up, the price per person, and the total revenue.
In skeet shooting, the clay pigeon is launched from a height of 4 feet and reaches a maximum height of 164 feet at a distance of 80 feet from the launch site.
Write an equation for the height of the clay pigeon in terms of the horizontal distance it has traveled.
The batter in a softball game hits the ball when it is 4 feet above the ground. The ball reaches the greatest height on its trajectory, 35 feet, directly above the head of the left-fielder, who is 200 feet from home plate.
Write an equation for the height of the softball in terms of its horizontal distance from home plate.
The rate at which an antigen precipitates during an antigen–antibody reaction depends on the amount of antigen present. For a fixed quantity of antibody, the time required for a particular antigen to precipitate is given in minutes by
\begin{equation*}
t=2w^2-20w+54
\end{equation*}
where \(w\) is the quantity of antigen present, in grams. For what quantity of antigen will the reaction proceed most rapidly, and how long will the precipitation take?
