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AveGali [126]
3 years ago
8

The surface area of a right circular cone of radius r and height h is S = πr√ r 2 + h 2 , and its volume is V = 1 3 πr2h. What i

s the largest volume of a right circular cone if its surface area is S? Use Lagrange multipliers to solve the problem.
Mathematics
1 answer:
kirill115 [55]3 years ago
3 0

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,

S=\pi r\sqrt{r^2+h^2}

and volume,

V=\frac{1}{3}\pi r^2 h

To find the largest volume if the surface area is S=8 (say), then applying Lagranges multipliers,

f(r,h)=\frac{1}{3}\pi r^2 h

subject to,

g(r,h)=\pi r\sqrt{r^2+h^2}=8\hfill (1)

We know for maximum volume r\neq 0. So let \lambda be the Lagranges multipliers be such that,

f_r=\lambda g_r

\implies \frac{2}{3}\pi r h=\lambda (\pi \sqrt{r^2+h^2}+\frac{\pi r^2}{\sqrt{r^2+h^2}})

\implies \frac{2}{3}r h= \lambda (\sqrt{r^2+h^2}+\frac{ r^2}{\sqrt{r^2+h^2}})\hfill (2)

And,

f_h=\lambda g_h

\implies \frac{1}{3}\pi r^2=\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}

\implies \lambda=\frac{r\sqrt{r^2+h^2}}{3h}\hfill (3)

Substitute (3) in (2) we get,

\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}})

\implies \frac{2}{3}rh=\frac{r}{3h}(2r^2+h^2)

\implies h^2=2r^2

Substitute this value in (1) we get,

\pi r\sqrt{h^2+r^2}=8

\implies \pi r \sqrt{2r^2+r^2}=8

\implies r=\sqrt{\frac{8}{\pi\sqrt{3}}}\equiv 1.21252

Then,

h=\sqrt{2}(1.21252)\equiv 1.71476

Hence largest volume,

V=\frac{1}{3}\times \pi \times\frac{\pi}{8\sqrt{3}}\times 1.71476=0.407114

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