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REALLY NEED HELP WITH THIS

REALLY NEED HELP WITH THIS-example-1

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well, profit equations are usually a parabolic path like a camel's hump, profit goes up up and reaches a maximum then back down, the issue is to settle at the maximum point, thus the maximum profit.

So for this equation, like any quadratic with a negative leading coefficient, the maximum will occur at its vertex, with x-price at y-profit.


\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-5}x^2\stackrel{\stackrel{b}{\downarrow }}{+209}x\stackrel{\stackrel{c}{\downarrow }}{-1090} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 209}{2(-5)}~~~~ ,~~~~ -1090-\cfrac{ (209)^2}{4(-5)}\right) \implies\left( - \cfrac{ 209 }{ -10 }~~,~~-1090 - \cfrac{ 43681 }{ -20 } \right)


\left( \cfrac{ 209 }{ 10 }~~,~~-1090 + \cfrac{ 43681 }{ 20 } \right)\implies \left( \cfrac{ 209 }{ 10 }~~,~~-1090 + 2184.05 \right) \\\\\\ ~\hfill~\stackrel{ \$price\qquad profit }{(~20.90~~,~~ 1094.05~)}~\hfill~

answered
User Yassin Hajaj
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