My profit formula is P = -.005x^2 + 4x - 200. The question wants me to find the number of donuts sold (what they are talking about in the problem) that will generate the highest profit. How do I determine that? Thank you.
the highest profit to be attained is 0,making it a quadratic equation,find x,any value not x gives a loss(-P)
highest profit is at the vertex, when x = -b/2a = 4/.01 = 400
P(400) = 600
Steve,
Thank you - this makes sense that the highest profit is at the vertex, but where is the x = =b/2a from, and where is the 4/.01 = 400 from? I'm having trouble connecting those numbers.
Thanks for clarifying for me.
Steve is probably off line now, so I will answer.
For any quadratic function of the form
y = ax^2 + bx + c
the x of the vertex is -b/(2a), once you have the x you just sub that back into the equation to get the y of the vertex.
In your case:
P = -.005x^2 + 4x - 200
a = -.005
b = 4
then x = -b/(2a) = -4/(-.010) = 4/.01 = 400
P(400) = -.005(400)^2 + 4(400) - 200 = 600
so the vertex is (400,600)
Thank you both!
To determine the number of donuts sold that will generate the highest profit, we need to find the x-value (number of donuts sold) where the profit function reaches its maximum point. This can be done using calculus.
The given profit formula is P = -.005x^2 + 4x - 200. To find the maximum point, we need to differentiate the profit function with respect to x (take the derivative), and then find the x-value that makes the derivative equal to zero.
Let's find the derivative of the profit function:
dP/dx = -0.005(2x) + 4
= -0.01x + 4
To find the critical point, we set the derivative equal to zero and solve for x:
-0.01x + 4 = 0
-0.01x = -4
x = -4 / -0.01
x = 400
So, when the number of donuts sold (x) is 400, the profit function reaches its maximum point.
Therefore, selling 400 donuts will generate the highest profit according to the given profit formula.