Respuesta :
The quadratic equation models the revenue with respect to the unit price is:
[tex]R(p)=-2p^2+4000p[/tex]a) To find the prices when the revenue is zero, first zero the equation:
[tex]-2p^2+4000p=0[/tex]Using the quadratic equation, where "p" is the independent variable, normally represented with the letter "x"
[tex]p=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Where
a is the coefficient of the quadratic term, in this case, a= -2
b is the coefficient of the x-term, in this case, b= 4000
c is the constant of the quadratic equation, in this case, c= 0
[tex]\begin{gathered} p=\frac{-4000\pm\sqrt[]{(4000)^2-4\cdot(-2)\cdot0}}{2(-2)} \\ p=\frac{-4000\pm\sqrt[]{16000000-0}}{-4} \\ p=\frac{-4000\pm\sqrt[]{16000000}}{-4} \\ p=\frac{-4000\pm4000}{-4} \end{gathered}[/tex]
Solve the sum and difference separately:
-Sum:
[tex]\begin{gathered} p=\frac{-4000+4000}{-4} \\ p=\frac{0}{-4} \\ p=0 \end{gathered}[/tex]-Difference
[tex]\begin{gathered} p=\frac{-4000-4000}{-4} \\ p=\frac{-8000}{-4} \\ p=2000 \end{gathered}[/tex]At the unit prices 0 and 2000, the revenue will be zero.
b) You have to find the values of p for which the revenue is equal to 400000, to do so, equal the quadratic equation to the given revenue value:
[tex]400000=-2p^2+4000p[/tex]Zero the equation
[tex]\begin{gathered} 400000-400000=-2p^2+4000p-400000 \\ 0=-2p^2+4000p-400000 \end{gathered}[/tex]Use the quadratic equation to determine the prices, use:
a=-2
b=4000
c=-400000
[tex]\begin{gathered} p=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ p=\frac{-4000\pm\sqrt[]{(4000)^2-4\cdot(-2)(-400000)}}{2\cdot(-2)} \\ p=\frac{-4000\pm\sqrt[]{16000000-3200000}}{-4} \\ p=\frac{-4000\pm\sqrt[]{12800000}}{-4} \\ p=\frac{-4000\pm3577.708764}{-4} \\ \end{gathered}[/tex]Sum:
[tex]\begin{gathered} p=\frac{-4000+3577.708764}{-4} \\ p=105.57 \end{gathered}[/tex]Difference:
[tex]\begin{gathered} p=\frac{-4000-3577.708764}{-4} \\ p=1894.427\approx1894.43 \end{gathered}[/tex]Between the prices $105.57 and $1894.43 the revenue will be greater than $400,000
