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]
1. 3/4
2. -3/4
1. 3/4
The profit function, p(x), can be modeled as a quadratic function of the form p(x) = ax^2 + bx + c. Since we know that the maximum profit occurs when Charlie sells 5 bracelets, we can substitute this value into the profit function to solve for a.
p(5) = a(5)^2 + b(5) + c
12 = 25a + 5b + c
0 = 25a + 5b + c - 12
-12 = 25a + 5b
Since the maximum profit occurs at the vertex of the parabola, we know that the x-coordinate of the vertex is given by the formula x = -b/2a. In this case, x = 5. By substituting the values of x and a, we can solve for b.
5 = -b / 2(3/4)
5 = -b / (3/2)
5 = -2b / 3
15 = -2b
b = -15/2
Now that we have found the values of a and b, we can construct Charlie's profit function:
p(x) = 3/4 x^2 - 15/2 x + c
To find the value of c, we can substitute the point (5, 12) into the function:
12 = 3/4(5)^2 - 15/2(5) + c
12 = 75/4 - 75/2 + c
12 = 75/4 - 150/4 + c
12 = -75/4 + c
12 + 75/4 = c
c = 63/4
Therefore, Charlie's profit function is:
p(x) = 3/4 x^2 - 15/2 x + 63/4