Given that Charlie will have a maximum profit of $12 when he sells 5 bracelets, we can create the following system of equations:
p(x) = ax^2 + bx + c
p(5) = 12
To find [a], [b], and [c], we can plug in the values from the second equation:
p(5) = a(5)^2 + b(5) + c = 12
25a + 5b + c = 12
Since we know that b = 0 (there is no linear term in the function since there is no specified revenue for each bracelet sold), we can simplify the equation to:
25a + c = 12
c = 12 - 25a
Next, we know that the profit function will have a maximum at x = h. Since the maximum profit is at x = 5, we can conclude that h = 5.
Therefore, now we have:
p(x) = a(x - 5)^2 + (12 - 25a)
To solve for [a], we can set the equation equal to the possibility of a max value of $12 when x = 5:
12 = a(5 - 5)^2 + 12 - 25a
12 = 12 - 25a
25a = 0
a = 0
So, [a] = 0
Thus, the profit function for Charlie's sales of beaded bracelets is:
p(x) = 0(x - 5)^2 + 12
p(x) = 12
3)
Charlie sells beaded bracelets as a side job. Charlie will have a maximum profit of $12 when he sells 5 bracelets.
Construct Charlie's profit function, p (x), to model his sales, in dollars, as a function of the number of bracelets he sells.
p(x) = [a] (x - [h])^2 + [k]
Find [a]
Find [h]
Find [k]
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