Answer:
Just see how are they alike and differents
Step-by-step explanation:
Here’s the correct answer and how I got the answer :)
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,
and volume,
![V=\frac{1}{3}\pi r^2 h](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%20r%5E2%20h)
To find the largest volume if the surface area is S=8 (say), then applying Lagranges multipliers,
subject to,
![g(r,h)=\pi r\sqrt{r^2+h^2}=8\hfill (1)](https://tex.z-dn.net/?f=g%28r%2Ch%29%3D%5Cpi%20r%5Csqrt%7Br%5E2%2Bh%5E2%7D%3D8%5Chfill%20%281%29)
We know for maximum volume
. So let
be the Lagranges multipliers be such that,
![f_r=\lambda g_r](https://tex.z-dn.net/?f=f_r%3D%5Clambda%20g_r)
![\implies \frac{2}{3}\pi r h=\lambda (\pi \sqrt{r^2+h^2}+\frac{\pi r^2}{\sqrt{r^2+h^2}})](https://tex.z-dn.net/?f=%5Cimplies%20%5Cfrac%7B2%7D%7B3%7D%5Cpi%20r%20h%3D%5Clambda%20%28%5Cpi%20%5Csqrt%7Br%5E2%2Bh%5E2%7D%2B%5Cfrac%7B%5Cpi%20r%5E2%7D%7B%5Csqrt%7Br%5E2%2Bh%5E2%7D%7D%29)
![\implies \frac{2}{3}r h= \lambda (\sqrt{r^2+h^2}+\frac{ r^2}{\sqrt{r^2+h^2}})\hfill (2)](https://tex.z-dn.net/?f=%5Cimplies%20%5Cfrac%7B2%7D%7B3%7Dr%20h%3D%20%5Clambda%20%28%5Csqrt%7Br%5E2%2Bh%5E2%7D%2B%5Cfrac%7B%20r%5E2%7D%7B%5Csqrt%7Br%5E2%2Bh%5E2%7D%7D%29%5Chfill%20%282%29)
And,
![f_h=\lambda g_h](https://tex.z-dn.net/?f=f_h%3D%5Clambda%20g_h)
![\implies \frac{1}{3}\pi r^2=\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}](https://tex.z-dn.net/?f=%5Cimplies%20%5Cfrac%7B1%7D%7B3%7D%5Cpi%20r%5E2%3D%5Clambda%20%5Cfrac%7B%5Cpi%20rh%7D%7B%5Csqrt%7Br%5E2%2Bh%5E2%7D%7D)
![\implies \lambda=\frac{r\sqrt{r^2+h^2}}{3h}\hfill (3)](https://tex.z-dn.net/?f=%5Cimplies%20%5Clambda%3D%5Cfrac%7Br%5Csqrt%7Br%5E2%2Bh%5E2%7D%7D%7B3h%7D%5Chfill%20%283%29)
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}})](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B3%7Drh%3D%5Cfrac%7Br%5Csqrt%7BR%5E2%2Bh%5E2%7D%7D%7B3h%7D%28%5Csqrt%7BR%5E2%2Bh%5E2%2B%7D%2B%5Cfrac%7Br%5E2%7D%7B%5Csqrt%7Br%5E2%2Bh%5E2%7D%7D%29)
![\implies \frac{2}{3}rh=\frac{r}{3h}(2r^2+h^2)](https://tex.z-dn.net/?f=%5Cimplies%20%5Cfrac%7B2%7D%7B3%7Drh%3D%5Cfrac%7Br%7D%7B3h%7D%282r%5E2%2Bh%5E2%29)
![\implies h^2=2r^2](https://tex.z-dn.net/?f=%5Cimplies%20h%5E2%3D2r%5E2)
Substitute this value in (1) we get,
![\pi r\sqrt{h^2+r^2}=8](https://tex.z-dn.net/?f=%5Cpi%20r%5Csqrt%7Bh%5E2%2Br%5E2%7D%3D8)
![\implies \pi r \sqrt{2r^2+r^2}=8](https://tex.z-dn.net/?f=%5Cimplies%20%5Cpi%20r%20%5Csqrt%7B2r%5E2%2Br%5E2%7D%3D8)
![\implies r=\sqrt{\frac{8}{\pi\sqrt{3}}}\equiv 1.21252](https://tex.z-dn.net/?f=%5Cimplies%20r%3D%5Csqrt%7B%5Cfrac%7B8%7D%7B%5Cpi%5Csqrt%7B3%7D%7D%7D%5Cequiv%201.21252)
Then,
![h=\sqrt{2}(1.21252)\equiv 1.71476](https://tex.z-dn.net/?f=h%3D%5Csqrt%7B2%7D%281.21252%29%5Cequiv%201.71476)
Hence largest volume,
![V=\frac{1}{3}\times \pi \times\frac{\pi}{8\sqrt{3}}\times 1.71476=0.407114](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cpi%20%5Ctimes%5Cfrac%7B%5Cpi%7D%7B8%5Csqrt%7B3%7D%7D%5Ctimes%201.71476%3D0.407114)
Answer:
a = 11.71 ; b = 15.56
Step-by-step explanation:
For this problem, we need two things. The law of sines, and the sum of the interior angles of a triangle.
The law of sines is simply:
sin(A)/a = sin(B)/b = sin(C)/c
And the sum of interior angles of a triangle is 180.
45 + 110 + <C = 180
<C = 25
We can find the sides by simply applying the law of sines.
length b
7/sin(25) = b/sin(110)
b = 7sin(110)/sin(25)
b = 15.56
length a
7/sin(25) = a/sin(45)
a = 7sin(45)/sin(25)
a = 11.71
Uhh my brains hurts looking at that