Asked 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=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.
Answers
Answered by
Damon
x^2 -56 x = -p - 300
complete the square to find vertex of parabola
x^2 -56 x + 784 = -p +484
(x-28)^2 = - (p-484)
so
vertex (max of p) is at (28,484)
complete the square to find vertex of parabola
x^2 -56 x + 784 = -p +484
(x-28)^2 = - (p-484)
so
vertex (max of p) is at (28,484)
Answered by
Aria
Damon,
Why does this explain the highest yeild?
It sounds like what you did was on a graph and you found the highest points
Why does this explain the highest yeild?
It sounds like what you did was on a graph and you found the highest points
Answered by
Linda
-4x^3 + 24x^2 – 24x + 4 by x^2 – 5x + 1
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.