Liu's Maximization

Let $\space y = (20x+3)^{\frac{1}{2}} + (20-3x)^{\frac{1}{2}}$ .

$\quad \Rightarrow \dfrac {dy}{dx} = \frac {1}{2} (20x+3)^{\frac{1}{2}-1} (20) + \frac {1}{2} (20- 3x)^{\frac{1}{2}-1} (-3)$

$\quad \quad \quad \quad = \dfrac {10} {\sqrt{20x+3}} - \dfrac {3} {2\sqrt{20-3x}} = \dfrac {20\sqrt{20-3x} - 3\sqrt{20x+3} } {2\sqrt{20x+3} \sqrt{20-3x}}$

We know that $y$ is maximum, when $\frac {dy}{dx} = 0$ , that is when:

$\quad 20\sqrt{20-3x} - 3\sqrt{20x+3} = 0\quad \Rightarrow 20\sqrt{20-3x} = 3\sqrt{20x+3}$

$\quad \Rightarrow 400(20-3x) = 9(20x+3) \quad \Rightarrow 8000 - 1200x = 180x + 27$

$\quad \Rightarrow 1380x = 7973 \quad \Rightarrow x = \dfrac {7973}{1380}$

$\quad \Rightarrow x = 5.777536232 \quad \Rightarrow \lfloor {10^4\dot{}x_0} \rfloor = \boxed{57775}$

Since we just featured the Cauchy Schwarz Inequality Wiki , let me point out an algebraic way of solving this.

$\begin{aligned} \sqrt{ 20x + 3 } + \sqrt{ 20 - 3x } & = \frac{ 1}{ \sqrt{3} } \sqrt{ 60x - 9 } + \frac{ 1}{ \sqrt{20} } \sqrt{ 400 - 60x } \\ & \leq ( \frac{1}{3} + \frac{1}{20} ) ( 60x - 9 + 400 - 60x ) \\ & = \frac{ 23}{60} \times 409 \\ \end{aligned}$

Equality occurs when:

$\frac{ \sqrt{ 60x-9} } { \frac{1}{ \sqrt{3} } } = \frac{ \sqrt{ 400-60x } } { \frac{1}{ \sqrt{20} } }$

which gives us $180x - 27 = 8000 - 1200 x \Rightarrow x = \frac{ 7973}{ 1380}$ .

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This approach easily generalizes and teaches us how to deal with $\sqrt{ f(x) } + \sqrt{ g(x) }$ where $f(x), g(x)$ are linear functions which have slopes of different signs.

$f'(x)=\frac{20}{\sqrt{2 \times (20x+3)}}-\frac{3}{\sqrt{2 \times (20-3x)}}$ $20\times\sqrt{2\times(20-3x)}-3\times\sqrt{2\times(20x+3)}=0$ $20\times\sqrt{2\times(20-3x)}=3\times\sqrt{2\times(20x+3)}$ $400\times(2\times(20-3x))=9(2*(20x+3))$ $400\times(20-3x)=9\times(20x+3)$ $8000-1200\times x=180\times x+27$ $(1200+180)x=8000-27$ $x_{0}=\frac{8000-27}{1200+180}$ $f_{MAX}=\sqrt{20 \times x_{0}+3}+\sqrt{20-3 \times x_{0}}$ $\Rightarrow \lfloor 10^4 \cdot x_{0} \rfloor = \boxed{57775}$