P(x)=-0.002x²+3.5x-1100.
Perhaps we can work out how many patterns they need to sell for maximum profit, and there are a few ways to do this. One way is to draw a graph, which would be a parabola in this case—an inverted U shape, where the hump of the U would tell us what the maximum profit would be. The graph would need to be accurately drawn to get an accurate answer.
We can use algebra to rewrite the expression for P(x) so that it’s easy to tell what value of x is needed to maximise the profit.
We can use calculus to find x which would maximise the profit. You would only do this if you knew how to use calculus.
We can use the formula to find the profit for any given value of x.
Let’s rewrite the formula algebraically.
-1100-(0.002x²-3.5x) is the same expression. We can take this further by completing the square:
-1100-0.002(x²-3.5x/0.002)=-1100-0.002(x²-3.5x/0.002+(3.5/0.004)²-(3.5/0.004)²).
Not very inspiring so far. But see:
-1100-0.002(x-3.5/0.004)²+0.002(3.5/0.004)².
-1100-0.002(x-3500/4)²+0.002(3500/4)². Still messy? Keep watching.
-1100-0.002(x-875)²+0.002(875)²=-1100-0.002(x-875)²+1531.25. Nearly there.
P(x)=431.25-0.002(x-875)². Got it!
P(x) is maximum when the squared term is zero because all other values will reduce 431.25. The squared term is zero when x=875, that is, when 875 patterns are sold. And the profit will be $431.25.