# Can You Maximize a Profit Function? Business calculus students know this problem all too well… how much of some thingamabob should I make to make the most money? I love me some dolla dolla bills as much as you do, so here’s how to maximize your profit.

Let’s say your cost in C dollars of making n thingamabobs is given by the function C(n)=100n+.05n2+150.  And let’s say we’re going to sell them for a hundred bucks a piece, so our revenue function is R(n)=100n.  Then our profit function is our revenue minus our cost:

P(n) = R(n) – C(n) = 100n – (100n+.05n2+150) = -.05n2+90n-150

or P(n) = -.05n2+90n-150.  The graph of our profit function looks like this: You see that big blue dot at the top?  That’s our maximum profit!  So how do we find out what n is at that point?  Well check this out.  You see this big green line that’s tangent to the function at the max? What do you think is the slope of that line? It’s zero! And, dear readers, what’s another way to say “slope of a tangent line”? The derivative of course! And it just so happens that where the derivative is equal to zero is exactly where we get our max profit.  So to find the number of thingamabobs we should make, we’ll take the derivative of P(n), set it equal to zero, then solve for n.

P(n) = -.05n2+90n-150

P'(n) = -.1n+90

-.1n+90 = 0

n = 900

This tells us we need to make 900 thingamabobs to get the most profit.  (Caution! Make sure the n you find is a max, not a min! The derivative is equal to zero at minimums too!)

Okay, to sum up.  To maximize profit, take the derivative of the profit function, set it equal to zero, then solve for your variable.  Now go forth Brightstormers and make some bank!  I accept royalties in the form of pints of ice cream.