Maximum value of equations #2

$P=\sqrt{x} - x$ .

$\Leftrightarrow P=\sqrt{x}-(\sqrt{x})^{2}$ .

$\Leftrightarrow P=-[(\sqrt{x})^{2} - \sqrt{x}+\frac{1}{4}]+\frac{1}{4}$

$\Leftrightarrow P=\frac{1}{4}-(\sqrt{x}-\frac{1}{2})^{2}$

Since $(\sqrt{x}-\frac{1}{2})^{2}$ must be equal or larger than 0 (because $x$ is a real number), $P$ must be equal or smaller than $\frac{1}{4}$ .

Therefore, the maximum value of $P$ is $\frac{1}{4}$ (when $x=\frac{1}{4}$ ).

So the answer is $1+4=\boxed{5}$

$\frac{\partial \left(\sqrt{x}-x\right)}{\partial x}\Rightarrow \frac{1}{2 \sqrt{x}}-1$ which equals 0 at $x=\frac14$ . A function reaches an extremum or saddle point when its derivative is 0. The second derivative $\frac{\partial \frac{\partial \left(\sqrt{x}-x\right)}{\partial x}}{\partial x}\Rightarrow -\frac{1}{4 x^{3/2}}$ is $-2$ at $x=\frac14$ . Therefore, this is a maximum.