The daily profit, P, for selling x units of a certain item at

Given: The daily profit, P, for selling x units of a certain item at a sporting goods store can be modeled by the function \(P(x) = -a(x -\frac{b}{2a})^2 + \frac{b^2}{4a}+c\)­, where a and b are positive constants and c is a nonnegative constant.

Asked: What is the maximum daily profit for selling this item?<br />

Since a is positive, the maximum value of \(-a(x -\frac{b}{2a})^2 = 0\) since -a is negative and the value of expression is 0 at \(x =\frac{b}{2a}\)<br />
Maximum profit P(x=b/2a) =\( \frac{b^2}{4a}+c\)­

(1) \(b^2 + 4ac = \frac{52ac}{3}\)
Maximum daily profit = \( \frac{b^2}{4a}+c = \frac{bˆ2 + 4ac }{ 4a} = \frac{52ac}{3*4a} = \frac{13c}{3}\)
Since the value of c is unknown
NOT SUFFICIENT

(2) c = 360
Maximum daily profit = \( \frac{b^2}{4a}+c = \frac{b^2}{4a} + 360\)­
Since value of \(\frac{b^2}{4a}\) is unknown
NOT SUFFICIENT

(1) + (2) 
(1) \(b^2 + 4ac = \frac{52ac}{3}\)
Maximum daily profit = \( \frac{b^2}{4a}+c = \frac{bˆ2 + 4ac }{ 4a} = \frac{52ac}{3*4a} = \frac{13c}{3}\)
(2) c = 360
Maximum daily profit \(= \frac{13c}{3} = \frac{13*360}{3} = 13*120 = 1560\)
SUFFICIENT

IMO C­