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 a function of the number of bracelets he sells.
p(x) = [a] (x-[h])^2 + [k]
Find [a]
To find the value of [a], we can use the given information that Charlie will have a maximum profit of $12 when he sells 5 bracelets. Since this profit is the maximum, we can set the derivative of the profit function equal to 0 to find the maximum point.
The general form of the profit function is given as p(x) = ax^2 + bx + c. The profit function can be simplified as p(x) = ax^2 + bx. Since the profit function is quadratic, we know that the profit function has a maximum point if the coefficient of x^2 is negative.
Given that the maximum profit is $12 when he sells 5 bracelets, we have the following data points: (5, 12)
Substitute the data point into the profit function:
12 = a(5)^2 + 5b
12 = 25a + 5b
Since we are interested in finding the value of [a], we have to eliminate b. We can find b by differentiating the profit function and setting it to 0:
p'(x) = 2ax + b
0 = 2a(5) + b
0 = 10a + b
b = -10a
Substitute b back into the equation:
12 = 25a - 5(10a)
12 = 25a - 50a
12 = -25a
a = -12/25
a = -0.48
Therefore, the value of [a] is -0.48.