Question

If x is the charge and R is the revenue, we could figure out that the relation between the charge and the revenue can be modeled by the function

R(x)=-4x^2+84x or equivalent R(x)=x(-4x+84)
Formula: If f(x)=ax^2+bx+c, then the maximum value occurs at ((-b)/2a, f ((-b)/2a))
1. Determine the charge that will maximize the revenue and the maximum amount of revenue. Show how you obtained your answer.
2. Determine the charge(s) that would result in ZERO revenue. SHow how you obtained your answer.

Answer 1

the maximum for R(x) which is a parabola opening downwards with a "hump", is at the top of the hump or namely its vertex, the vertex is where the maximum for R(x) is at the "x" charge value

3)

`\textit{vertex of a vertical parabola, using coefficients} \\\\ R(x)=\stackrel{\stackrel{a}{\downarrow }}{-4}x^2\stackrel{\stackrel{b}{\downarrow }}{+84}x\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)`

`\left(-\cfrac{ 84}{2(-4)}~~~~ ,~~~~ 0-\cfrac{ (84)^2}{4(-4)}\right) \implies \left( - \cfrac{ 84 }{ -8 }~~,~~0 - \cfrac{ 7056 }{ -16 } \right) \\\\\\ \left( \cfrac{ -21 }{ -2 } ~~~~ ,~~~~ 0 +441 \right)\implies\hspace{5em} \stackrel{\textit{max of R(x)}\quad at~`

4)

well, we can find that by simply setting R(x) = 0

`\stackrel{R(x)}{0}=-4x^2+84x\implies 0=-4x(x-21)\implies \cfrac{0}{-4}=x(x-21) \\\\\\ 0=x(x-21)\implies x= \begin{cases} 0\\ 21 \end{cases}`

well, clearly if you charge nothing, well Revenue will be Zilch, nothing, however the same parabola that is at 0 when x = 0, is also at 0 when x = 21, meaning, if you charge more than 21 bucks, folks won't want to pay for it, so revenue will go kaput! again.