Minimum-maximum problem with graph plotting

[Karol]

Homework Statement

The first derivative of the area $~\displaystyle A(a)=a\sqrt{r^2-\frac{a^2}{4}}$ is positive everywhere

Homework Equations

When f'(x)>0 → the function rises

The Attempt at a Solution

$$A'=a\frac{1}{2}\left( r^2-\frac{a^2}{4} \right)^{-1/2}\cdot\left( \frac{1}{4} \right)2a+\sqrt{r^2-\frac{a^2}{4}}$$
$$A'=...=\frac{2r^2-a^2}{\sqrt{4r^2-a^2}}>0$$
I found A'=0 at $~\displaystyle a=\frac{2}{\sqrt{5}}r$

 

[Ray Vickson]

Homework Statement

View attachment 214283
The first derivative of the area $~\displaystyle A(a)=a\sqrt{r^2-\frac{a^2}{4}}$ is positive everywhere

Homework Equations

When f'(x)>0 → the function rises

The Attempt at a Solution

$$A'=a\frac{1}{2}\left( r^2-\frac{a^2}{4} \right)^{-1/2}\cdot\left( \frac{1}{4} \right)2a+\sqrt{r^2-\frac{a^2}{4}}$$
$$A'=...=\frac{2r^2-a^2}{\sqrt{4r^2-a^2}}>0$$
I found A'=0 at $~\displaystyle a=\frac{2}{\sqrt{5}}r$

No: $A^{\prime} \neq 0$ at $a = r \; 2 /\sqrt{5},$ as you can see from your own expression for $A^{\prime}$.

Anyway, for a rectangle of base $a$ we need $a < 2r$, so having a maximum of $A(a)$ at $a > r$ is perfectly OK.

 

[Karol]

$$\frac{2r^2-a^2}{\sqrt{4r^2-a^2}}=0\rightarrow a=\sqrt{2}r$$
But still A'>0 for 0<a<2r
It should be negative for a>√2⋅r

 

[Ray Vickson]

$$\frac{2r^2-a^2}{\sqrt{4r^2-a^2}}=0\rightarrow a=\sqrt{2}r$$
But still A'>0 for 0<a<2r
It should be negative for a>√2⋅r

It IS negative for $a > r \sqrt{2}$.

If you don't believe it, try plotting $A(a)$ for $r = 1$ and $0 < a < 2$.

 

[Karol]

Thank you Ray!