The number that exceeds its square by

Solution:

Let the number be x.

The square of the number is $x^{2}$.

Let $f(x)=x-x^{2}$. Now, we need to find the value of $x$ for which $f(x)$ is maximum.

$f(x)=x-x^{2}$

Differentiating both sides with respect to x, we get

$f^{\prime}(x)=1-2 x$

For maxima or minima,

$f^{\prime}(x)=0$

$\Rightarrow 1-2 x=0$

 

$\Rightarrow x=\frac{1}{2}$

Now,

$f^{\prime \prime}(x)=-2<0$

So, $x=\frac{1}{2}$ is the point of local maximum of $f(x)$. Therefore, $f(x)$ is maximum when $x=\frac{1}{2}$.

Thus, the number that exceeds its square by the greatest amount is $x=\frac{1}{2}$.

The number that exceeds its square by the greatest amount is  $\frac{1}{2}$