Given:
price p = $25
number of x units = -8p + 400
Find: solve for the value of x
Solution:
To determine the value of x, simply replace the variable "p" in the equation with 25.
[tex]x=-8(25)+400[/tex]
Then, solve.
Multiply -8 and 25.
[tex]x=-200+400[/tex]
Add -200 and 400.
[tex]x=200[/tex]
Therefore, at $25, 200 units were sold.
Part F:
Since the equation of the revenue is R(x) = -8p² + 400p, then the graph must be a parabola opening down. Out of the 4 options, Only Options A and D show this.
However, upon comparing the two options, we see that the y-axis of the two graphs is different. Option A says that the y-axis is the price p while Option D says that the y-axis is the revenue R.
Since the given function is the revenue, then the y-axis should be R. The correct graph is Graph D.
Part G:
If the revenue is at least $4032, then let's replace the r(x) function with 4, 032.
[tex]4,032=-8p^2+400p[/tex]
To solve for p, let's equate the function to zero by subtracting both sides of the function by 4, 032.
[tex]\begin{gathered} 4,032-4032=-8p^2+400p-4032 \\ 0=-8p^2+400p-4032 \end{gathered}[/tex]
Then, let's solve for the value of "p" using the quadratic formula.
[tex]p=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Note that in the revenue function above, a = -8, b = 400, and c = -4032. Let's plug these values into the formula above.
[tex]p=\frac{-400\pm\sqrt{400^2-4(-8)(-4032)}}{2(-8)}[/tex]
Then, simplify.
[tex]p=\frac{-400\pm\sqrt{160,000-129,024}}{-16}[/tex][tex]p=\frac{-400\pm\sqrt{30,976}}{-16}[/tex][tex]p=\frac{-400\pm176}{-16}[/tex]
Separate the plus and minus operations.
[tex]p=\frac{-400+176}{-16}=\frac{-224}{-16}=14[/tex][tex]p=\frac{-400-176}{-16}=\frac{-576}{-16}=36[/tex]
The values of p are 14 and 36.
Hence, the company should charge a price between a minimum of $14 and a maximum of $36 to have a revenue of at least $4,032.
