David makes and sells chairs. The function p(x) = -10x^2 + 100x - 210, graphed below, indicates how much profit he makes in a month if he sells the chairs for 10 - x dollars each. What should David charge per chair to make the maximum profit, and what is the maximum profit he can make in a month?
I apologize, I am unable to show the graph.
I really need help. I appreciate anyone who is able to help me with this problem.
Oh, whoops my apologizes. I realized this question was already answered.
-10x^2 + 100x - 210 = 10(-x^2+10x-21) = -10(x-4)(x-7)
the vertex is midway between the roots, so
p(5) = 40
Thank you so very much for this answer! Have a blessed one!
No problem! I can still help you without the graph.
To find the price per chair that would maximize David's profit, we need to find the vertex of the quadratic function p(x) = -10x^2 + 100x - 210. The vertex represents the maximum point on the graph, which corresponds to the maximum profit.
The formula for the x-coordinate of the vertex of a quadratic function in the form ax^2 + bx + c is given by x = -b / (2a).
In this case, a = -10, b = 100, and c = -210. Plugging these values into the formula, we find:
x = -100 / (2 * -10)
x = -100 / -20
x = 5
So, David should charge $5 per chair to maximize his profit.
To find the maximum profit, we substitute the value of x (5) back into the original function p(x):
p(5) = -10(5)^2 + 100(5) - 210
p(5) = -10(25) + 500 - 210
p(5) = -250 + 500 - 210
p(5) = 40
Therefore, the maximum profit David can make in a month is $40.
To summarize:
- David should charge $5 per chair to make the maximum profit.
- The maximum profit he can make in a month is $40.