Question

Uppose that the manufacturer of a gas clothes dryer has found​ that, when the unit price is p​ dollars, the revenue R​ (in dollars) is . What unit price should be established for the dryer to maximize​ revenue? What is the maximum​ revenue?


`R(p)= -7px^(2) +21,000p`

The unit price that should be established to maximize revenue is ​$

enter your response here

Answer 1

1500090000 and 9990aa

Answer 2

Check the picture below, that's a parabolic path for an object for a certain initiali velocity. anyhow, so long the quadratic equation has a negative leading coefficient, we'd get a "hump" and at its peak or vertex, is its maximum.

So in short, R(p) in this case is a parabolic path and its maximum is at its vertex, let's find it.

`\textit{vertex of a vertical parabola, using coefficients} \\\\ R(p)=\stackrel{\stackrel{a}{\downarrow }}{-7}p^2\stackrel{\stackrel{b}{\downarrow }}{+21000}p\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 21000}{2(-7)}~~~~ ,~~~~ 0-\cfrac{ (21000)^2}{4(-7)}\right) \implies\left( - \cfrac{ 21000 }{ -14 }~~,~~0 - \cfrac{ 441000000 }{ -28 } \right)`

`\left( \cfrac{ 21000 }{ 14 }~~,~~0 + \cfrac{ 441000000 }{ 28 } \right)\implies \left( \cfrac{ 21000 }{ 14 }~~,~~0 + 15750000 \right) \\\\\\ ~\hfill~ {\Large \begin{array}{llll} (~\stackrel{ \textit{unit price} }{1500}~~,~~ \stackrel{ \textit{maximum revenue} }{15750000}~) \end{array}}~\hfill~`