Given the volume of a grain silo and the height of the cylinder of the silo, solve for the radius.

Question

Q: A grain silo consists of a cylindrical main section and a hemispherical roof. If the total volume of the silo (including the part inside the roof section) is 16,000 ft3 and the cylindrical part is 30 ft tall, what is the radius of the silo, correct to the nearest tenth of a foot?

I started working from this equation for volume (cylinder and half of a sphere)...

V=(pi)(r^2)(h)+(2(pi)r^3)/(3)

Plugged in the given values and solved for the radius but haven't gotten the correct answer yet.

Can anyone help?

Answer 1

As Parcly Taxel answered, we know that $r=10$ is an underestimation. Use Newton method using $r_0=10$ for finding the zero of function $$f(r)=\pi r^2\left(\frac23 r+30\right)-16000 $$ The iterates will be given by $$r_{n+1}=\frac{2 \pi r_n^3+45 \pi r_n^2+24000}{3 \pi r_n(r_n+30)}$$ and will be $$r_1=11.7829$$ $$r_2=11.6176$$ $$r_3=11.6161$$ which is the solution for six significant figures.

As Parcly Taxel commented, using Cardano's method would give a monster as a formula.

Answer 2

$$V=h\pi r^2+\frac23\pi r^3=16000$$ $$30\pi r^2+\frac23\pi r^3=16000$$ $$\pi r^2\left(\frac23 r+30\right)=16000$$ This is a cubic equation in $r$. If you want to avoid the tedious Cardano's formula for solving such equations, you need numerical methods, which are relevant here because the solution only needs to be accurate to a tenth of a place.

Say we guess that $r=10$ feet, which gives us a volume of 11519 cubic feet. If we try $r=12$ we get 17191 cubic feet. Since these two values are on opposite sides of the required 16000, we can try values between these two limits to find $r$.

• $r=11.5$ gives 15650 cubic feet, less than 16000.
• $r=11.7$ gives 16256 cubic feet.
• $r=11.6$ gives 15951 cubic feet.
• $r=11.65$ gives 16103 cubic feet.

These last two evaluations give $11.6<r<11.65$, i.e. the radius of the silo is 11.6 feet to one decimal place. (The exact answer is $11.6161\dots$ feet.)