Algebra
posted by Aria .
If the profit equation for the monthly sales of tile sets is P = x^2 + 56x  300, how do I find the number of tiles that yeilds the most profit.
x=number of tile sets. The amount of money for costs each month is set at $300.
I saw an answer as 28 tile sets would yeild the highest profit but I don't understand why that is the answer, if it is.
Could you help me? Thanks so much.

x^2 56 x = p  300
complete the square to find vertex of parabola
x^2 56 x + 784 = p +484
(x28)^2 =  (p484)
so
vertex (max of p) is at (28,484) 
Damon,
Why does this explain the highest yeild?
It sounds like what you did was on a graph and you found the highest points 
4x^3 + 24x^2 – 24x + 4 by x^2 – 5x + 1
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