Find the number of units that maximize revenue

Question

Find the number of units that maximize revenue

R=-.01x²+400x

R= total revenue

x= units sold

Answer 1

R(x) = -.01x² + 400x is a parabola that is upside down (leading negative) so it will have a maximum at the vertex. The x coordinate of the vertex is -b/(2a) = -400/[2(-.01)] = 20,000. So that is the number of units that produce the maximum revenue.

Answer 2

Derivative = slope. At a max or min, slope is zero right? Just imagine a maximum point on the graph. So take the derivative: R'(x) = -.02x + 400.

Set R'(x) = 0; add over the -.02x and divide by .02 (just algebra) and get R'(x) = 0 when x = 20,000.

This tells you the slope is zero so it could be a maximum or minimum. Check to see if R'(x) changes from negative to positive at x = 20,000 (it's zero there so it was either negative and going up to positive or it was positive and going down to negative).

You could also take the second derivative which is R''(x) = -0.2x; this tells you that the derivative(slope) of the slope is negative so the slope is decreasing constantly. Think if the slope is 1 and then decreases to .5 and then to 0 and then to -.5 and then -1, that would look like a curve faced down right? So there would be a maximum, AKA if the 2nd derivative is negative the graph is concave down and you get a maximum.

Since we confirmed it's a maximum, we can say profit is maximized at x = 20,000 units. Please don't hesitate to comment if you would like a more in-depth explanation.