Question

Zadanie optymalizacyjne z matematyki. Proszę o rozwiązanie i wyjaśnienie

Answer 1

Volume of the pyramid:

$V=\dfrac{s^2h}3$

Perimeter of the cross-section:

$40=\sqrt2\,s+2\sqrt{\dfrac{s^2}2+h^2}=\sqrt2\left(s+\sqrt{s^2+2h^2}\right)$

$\implies h=\sqrt{\dfrac{(20\sqrt2-s)^2-s^2}2}=\sqrt{400-20\sqrt2\,s}$

Area of the cross-section:

$P=\dfrac12(\sqrt2\,s)h=\dfrac{sh}{\sqrt2}$

$\implies P=\dfrac{s\sqrt{400-20\sqrt2\,s}}{\sqrt2}=s\sqrt{200-10\sqrt2\,s}$

First derivative test:

$\dfrac{\mathrm dP}{\mathrm ds}=\dfrac{20\sqrt2-3s}{\sqrt{4-\dfrac{\sqrt2}5s}}=0\implies s=\dfrac{20\sqrt2}3$

Then the height of the cross-section/pyramid is

$h=\sqrt{400-20\sqrt2\,s}=\dfrac{20}{\sqrt3}$

The volume of the pyramid that maximizes the cross-sectional area $P$ is

$V=\dfrac{\left(\frac{20\sqrt2}3\right)^2\frac{20}{\sqrt3}}3=\dfrac{16000}{27\sqrt3}$