Find Maximum of this Function

Question

I'm sure this is a simple solution, but I can't find a proper explanation online.

I'm currently reading Algorithmic Game Theory, and at one point the author states that given the function

$x(1-t-x)$, the highest value $x$ can be is $ \frac{(1-t)}{2} $.

I can see this is true from just plugging in values of $t$ and finding the critical points. For example $x(1-0.5-x)$, $x$ has a value of $.25$

Can someone please explain to me how the author deduced $\frac{(1-t)}{2}$?

Thanks in advance!

Answer 1

This can be done without calculus. Just completing the square in algebra-precalculus will do.

\begin{align} x(1-t-x) &= -(x^2 - (1-t)x) \\ &= -\left(x^2-2\cdot\left(\frac{1-t}{2}\right) x + \left(\frac{1-t}{2}\right)^2\right) + \left(\frac{1-t}{2}\right)^2 \\ &= \underbrace{-\left((x-\frac{1-t}{2}\right)^2}_{\le 0} + \left(\frac{1-t}{2}\right)^2 \end{align}

Therefore, $x = \dfrac{1-t}{2}$ maximizes $x(1-t-x)$.

Answer 2

write your term in the form $$-\left(x^2-2\frac{1-t}{2}x+\left(\frac{1-t}{2}\right)^2\right)+\left(\frac{1-t}{2}\right)^2$$

Answer 3

Expand: $$x(1-t-x)=-x^2+(1-t)x.$$ Note that the inverted parabola attains max at its vertex: $$x_0=-\frac{b}{2a}=-\frac{1-t}{-2}=\frac{1-t}{2}.$$

Answer 4

Here a simple geometric argument:

The graph of $p(x) = x(1-t-x)$ is a parabola open downwards which intersects the $x$-axis at $x=0$ and $x= 1-t$.

So, the maximum is attained at the vertex which is because of the symmetry of the parabola exactly in the middle between the two zeros:

$$x_{max} = \frac{0+1-t}{2}= \frac{1-t}{2} \rightarrow y_{max} = \frac{1-t}{2}(1-t - \frac{1-t}{2}) = \left( \frac{1-t}{2} \right)^2$$