Respuesta :
Answer:
Required largest volume is 0.407114 unit.
Step-by-step explanation:
Given surface area of a right circular cone of radious r and height h is,
[tex]S=\pi r\sqrt{r^2+h^2}[/tex]
and volume,
[tex]V=\frac{1}{3}\pi r^2 h[/tex]
To find the largest volume if the surface area is S=8 (say), then applying Lagranges multipliers,
[tex]f(r,h)=\frac{1}{3}\pi r^2 h[/tex]
subject to,
[tex]g(r,h)=\pi r\sqrt{r^2+h^2}=8\hfill (1)[/tex]
We know for maximum volume [tex]r\neq 0[/tex]. So let [tex]\lambda[/tex] be the Lagranges multipliers be such that,
[tex]f_r=\lambda g_r[/tex]
[tex]\implies \frac{2}{3}\pi r h=\lambda (\pi \sqrt{r^2+h^2}+\frac{\pi r^2}{\sqrt{r^2+h^2}})[/tex]
[tex]\implies \frac{2}{3}r h= \lambda (\sqrt{r^2+h^2}+\frac{ r^2}{\sqrt{r^2+h^2}})\hfill (2)[/tex]
And,
[tex]f_h=\lambda g_h[/tex]
[tex]\implies \frac{1}{3}\pi r^2=\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}[/tex]
[tex]\implies \lambda=\frac{r\sqrt{r^2+h^2}}{3h}\hfill (3)[/tex]
Substitute (3) in (2) we get,
[tex]\frac{2}{3}rh=\frac{r\sqrt{R^2+h^2}}{3h}(\sqrt{R^2+h^2+}+\frac{r^2}{\sqrt{r^2+h^2}})[/tex]
[tex]\implies \frac{2}{3}rh=\frac{r}{3h}(2r^2+h^2)[/tex]
[tex]\implies h^2=2r^2[/tex]
Substitute this value in (1) we get,
[tex]\pi r\sqrt{h^2+r^2}=8[/tex]
[tex]\implies \pi r \sqrt{2r^2+r^2}=8[/tex]
[tex]\implies r=\sqrt{\frac{8}{\pi\sqrt{3}}}\equiv 1.21252[/tex]
Then,
[tex]h=\sqrt{2}(1.21252)\equiv 1.71476[/tex]
Hence largest volume,
[tex]V=\frac{1}{3}\times \pi \times\frac{\pi}{8\sqrt{3}}\times 1.71476=0.407114[/tex]
