Maximums on Quadratic Functions

Question

How do you find the maximum of a quadratic function? Specifically, $R(x) = -4x^2 + 4000x$

Answer 1

I assume from your question that you have not had differential calculus, or that this question is posed in a calculus course before coming to the concept of derivatives, which would make it easier.

You find the maximum by the trick of completing the square:

$$R(x) = -4x^2 + 4000x = -4 (x^2 - 1000x) = -4[(x-500)^2-500^2] \\= -4(x-500)^2+1000000$$

The $-4(x-500)^2$ is never positive, and if it is non-zero, that can only make the value of $R(x)$ smaller. So the maximum is attained when $x = 500$, and there, the value is ] $$ R(500) = 1000000 $$

Answer 2

Find the roots of $$-4x^2 + 4000x=0$$ These are $x=0$ and $x=1000$. The maximum of the quadratic function appears at the midpoint between the roots i.e. at $x=500$ and its value is $$R(500)=-4*500^2+4000*500=1000000$$