Two numbers add to make 1000. What would they have to be to maximise their product?
We can write the two numbers as x and 1000-x, because x+(1000-x) = 1000 for any x we pick. Now we want to maximise the product x(1000-x); expanding the brackets gives us 1000x-x2 which is easier to work with for us. I'm going to call this f(x).To maximise a function, you need to find its critical points and then check whether they are maxima or minima. Remember that critical points are points where the gradient is 0 (ie when the derivative is 0), and you can check the nature of a critical point by finding the sign of the second derivative. This means we can use the gradient function dy/dx to find our values for x.Let's differentiate f(x):
dy/dx = 1000-2xIf dy/dx = 0, then 1000-2x = 0So 2x = 1000, ie x = 500 and this is the only critical point.Checking the second derivative:d2y/dx2 = -2 < 0, so the point is a maximum.
Finally we check that the answer makes sense: 500(1000-500) = 5002 = 250000 which is very large, so our answer looks correct.You could also do this problem by completing the square, or by drawing the graph of f(x) = x(1000-x) and finding the maximum point. Try it!
