Question 6296: Hello
I am trying to solve the following:
A company has a profit function: P(n) = -3n2 + 36n + 80 to model the profit (in $1000) in selling n units. How many units must be sold to maximize the profit?
I am stuck as to how to begin. To maximize profits you would want an infinite number of units sold...??? What do I set my = to?
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Your observation that selling an infinite number of units would maximise profits does not fit the given profit function: P(n) = -3n^2 + 36n + 80 which is a quadratic equation.
You will notice that the graph of the profit function is a parabola (not a straight line) and you'll also notice that this parabola opens downward (the coefficient of n^2 is negative) so this means that the function has a maximum.
Quadratic equations have exactly one maximum (if they open downward) or one minimum (if they open upward). So, to answer the question, you'll need to find the value of selling units (n) at the maximum point on the parabola. This is also known as the vertex of the parabola.
The x-coordinate (or in this case, n) of the vertex of a parabola is located at x = -b/2a
You will need to make sure that your quadratic is in the "standard" form:
ax^2 + bx + c
In this problem, a = -3, b = 36, and c = 80.
So, you can find the vertex (maximum) of this parabola by substituting the appropriate values of a and b from the given profit function.
n = -36/2(-3)
n = -36/-6
n = 6 This is the number of units that must be sold in order to maximise profits.
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