A farmer is building a new cylindrical silo with a flat roof and an earthen floor that will hold 20000 m^3 of corn. What dimensions of the silo will minimize the materials required for construction?

2 Answers
Jun 21, 2018

I tried this:

Explanation:

Have a look:
enter image source here

Jun 21, 2018

Numerically, to three significant figures, r=h=18.5 m.

Explanation:

The question asks us to minimise the surface area of the cylinder ignoring the floor.

If the radius of the silo is r, then the area of the circular roof is pir^2. The area of the circular wall is its height times its circumference: 2pirh. So the surface area of the construction is 2pirh+pir^2=pir(2h+r).

The volume of the silo is given as 20000 m^3, and is the area of the base times the silo's height: pir^2h. If we set this to the given amount, we can derive an expression for h in terms of r that we can substitute in to our surface area expression:

pir^2h=20000
h=20000/(pir^2)

So the surface area is
pir(2*20000/(pir^2)+r)=40000/r+pir^2.

As this is now a function only of r, we can find its extrema.
Let A(r)=40000/r+pir^2. Then (dA)/(dr)=-40000/r^2+2pir.

(dA)/(dr)=0rArr-40000/r^2+2pir=0rArr2pir^3=40000
r^3=20000/pirArrr=root(3)(20000/pi)=10root(3)(20/pi)

So the function has a single real root, the other two being complex conjugates of each other. Now let us deduce its nature via inspection of the second derivative.

(d^2A)/(dr^2)=80000/r^3+2pi

(d^2A)/(dr^2)(10root(3)(20/pi))=80000pi/20000+2pi=4pi+2pi=6pi>0

This the construction surface area is minimum at the single extremum.

Thus the desired dimensions are:
Radius, r=10root(3)(20/pi)
Height, h=20000/(pir^2)=20000/(pi*100root(3)(400/pi^2))=200/root(3)(400pi)=100/root(3)(50pi)=10root(3)(20/pi), which is the same expression as for the radius. So the needed height and the needed radius are the same.

Numerically, to three significant figures, r=h=18.5 m.

Sanity check the solution by plotting out the graph of radius vs. surface area:
graph{y=40000/x + pi x^2 [10, 30, 3000, 4500]}