Answer:
Step-by-step explanation:
You can use calculus to derive the equation for the volume of a cone.
if we take an circle at the tip of the cone the radius is 0 but what we want to do is integrate along the x axis from 0 up to the height of the cone H.
So, the equation becomes
[tex]\int\limits^H_0 {\pi r^{2} } \, dx[/tex]
r is related to x by the formula for a line; r = m*x+b
m is the rise over the run which is given by the maximum height H divided by the maximum radius R. So m = H/R and the intercept b = 0
now we have a formula for r in terms of x and with a little variable manipulation we can get a formula for x in terms of r
r = (R/H)*x
If we replace r by this formula in our integral we get:
[tex]\int\limits^H_0 {\pi ((R/H)*x)^{2} } \, dx[/tex]
since [tex]\pi[/tex], R, and H are constants we can move them outside the integral:
[tex]\pi (R^{2} /H^{2} )*\int\limits^H_0 {x^{2} } \, dx[/tex]
the integral is easy. It is [tex]\frac{x^{3}}{3}\left \{{{x=H}\atop{x=0}} \right.[/tex] which is equal to [tex]H^{3} /3[/tex]
substituting back into the original formula we get the equation for the volume of a cone.
V = [tex](\pi R^{2}H)/3[/tex]
Use this formula with the values given in the question to solve for R.
183.17 = [tex]\pi[/tex][tex]R^{2}[/tex]*7/3
rearranging we get
[tex]\sqrt{183.17*3/(7\pi ) }[/tex] = R
R ≅ 5
