Maximize volume of pyramid

Let $a$ be the side of the base, $H$ height of the pyramid, and $h$ height of the side of the pyramid. Then area of the pyramid is:

$A=a^{2}+2ah$

Substituting $h=\sqrt{\frac{a^{2}}{4}+H^{2}}$ , we get:

$A=a^{2}+2a\sqrt{\frac{a^{2}}{4}+H^{2}}=a^{2}+a\sqrt{a^{2}+4H^{2}}$

Factoring and rearranging:

$A=a(a+\sqrt{a^{2}+4H^{2}})$

$a=\frac{A}{a+\sqrt{a^{2}+4H^{2}}}$

Squaring both sides:

$a^{2}=\frac{A^{2}}{a^{2}+2a\sqrt{a^{2}+4H^{2}}+a^{2} + 4H^{2}}=\frac{A^{2}}{2a^{2}+2a\sqrt{a^{2}+4H^{2}}+ 4H^{2}}=\frac{A^{2}}{2A+ 4H^{2}}$ .

Now, the formula $V=\frac{a^{2}H}{3}$ becomes:

$V=\frac{A^{2}H}{3(2A+4H^{2})}$

Differentiate for $H$ :

$\frac{\partial V}{\partial H}=\frac{\partial }{\partial H} (\frac{A^{2}H}{3(2A+4H^{2})})$

Since we are looking for the maximum volume, the derivative will be equal to $0$ . By the quotient rule:

$0=\frac{\partial A^{2}H}{\partial H} \times (6A+12H^{2})-\frac{\partial (6A+12H^{2})}{\partial H}\times A^{2}H$

$0=A^{2} \times (6A+12H^{2})-24H\times A^{2}H$

$6A^{3}+12A^{2}H^{2}=24A^{2}H^{2}$

$6A^{3}=12A^{2}H^{2}$

$A=2H^{2}\Rightarrow H^{2}=\frac{{A}}{2}\Rightarrow H=\sqrt{\frac{{A}}{2}}$

Now plugging this value in formula for $a$ :

$a^{2}=\frac{A^{2}}{2A+4\times\frac{2A}{4}}=\frac{A^{2}}{4A}\Rightarrow a=\frac{1}{2}\sqrt{A}$

Finally:

$V_{max}=\frac{\frac{A}{4}\times \sqrt{\frac{A}{2}}}{3}$

$V_{max}=\frac{A\sqrt{A}}{12\sqrt{2}}$ .