ok... so, we know the price is
p = -11x + 231
a)
how much should be charged with only 70 sets? 70 is the quantity or "x"
thus the price should be p = -11(70) + 231 <-- because x = 70
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b)
first off, we need the Revenue function, or R(x)
revenue is flatly, how much you sold, at what price
we know the quantity is "x", we know the price is -11x+231
so R(x) is x*p or
R(x) = -11x²+231x
what quantity will maximize it? well, notice, is a quadratic with a negative leading term's coefficient, notice the picture below, anyhow, the graph uses Profit or P(x), but is the same case for R(x), they're both parabolic graphs opening downwards, and their vertex is the highest point, that is, maximum profit or revenue
so... what's the vertex of R(x) = -11x²+231x ?
well [tex]\bf \begin{array}{lccclll}
R(x)=&-11x^2&+231x&+0\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)[/tex]
so, the revenue is the maximum at [tex]\bf c-\cfrac{b^2}{4a}\implies 0-\cfrac{231^2}{4(-11)}\implies \cfrac{53361}{44}\implies \cfrac{4851}{4}\implies 1212.75[/tex]
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c)
what is "p" at that vertex? namely, what's the price at that maximum revenue? well, is [tex]\bf -\cfrac{b}{2a}\implies -\cfrac{231}{2(-11)}\implies \cfrac{-231}{-22}\implies \cfrac{21}{2}\implies 10.5[/tex]
