Much ado about a roof!

Geometry Level 5

A medieval ruler once wanted to build a magnificent roof over his new apartments. He contacted his architects and they finally came up with two possibilities. One was a simple hemispherical roof of radius R R . The other was a complex idea which was arrived at by first taking a right circular cone of radius and height R R and making cuts parallel to the base at regular intervals of height R n \frac{R}{n} . This gives us n n pieces of equal height. Between each two pieces they want to insert a cylinder of same height R n \frac{R}{n} and radius equal to that of the surface on which it is kept.

The king listens to all of this and summons the court mathematician and asks him to point out an economically more feasible project.

The mathematician knows that the cost of cementing a structure is some C m − 3 C m^{-3} and cost of painting an unit area of the hemisphere was twice that of painting an unit area of the more complicated structure as the painter had to arrange for additional support and balance for the former. The bookish mathematician, however calculates assuming that the hemispherical and complicated roof, both are solid and not hollow!

According to the mathematician, what is the right choice for the cash strapped king?

Take R = 1 m R=1\text{ m} and n = 50 n=50 .

The mathematician is unable to decide Either will do Cone and cylinder roof Hemispherical roof

Rohith M.Athreya by Rohith M.Athreya , 22, India

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1 solution

There are in all 50 pieces of cone and 49 of cylinder.
From bottom, a cone piece can be seen to has more volume then cylinder piece above it; so also the surface area of cone piece is more than the cylinder piece above it. Over and above, the last cone piece has no cylinder piece above it. So the volume and surface of the cylinder pieces are less than the volume and areas of the cone pieces. V o l u m e o f h e m i s p h e r e = 1 2 ∗ 4 ∗ π ∗ R 3 3 = 2 3 ∗ π ∗ R 3 = V o l u m e o f T W O c o n e s . H e m i s p h e r e a n d c o n e w i t h s a m e r a d i u s a n d h e i g h t h a v e t h e s a m e s u r f a c e a r e a . B u t H e m i s p h e r e c o a s t i s t w i c e f o r C o n e − c y l i n d e r . S o H e m i s p h e r e c o a s t i s = T W O c o n e c o s t . B u t C o n e − c y l i n d e r c o s t i s l e s s t h a n T w o c o n e c o s t , a n d s o l e s s t h a n h e m i s p h e r e c o s t . Volume ~of~ hemisphere=\frac1 2*\dfrac { 4 *\pi*R^3} 3=\frac 2 3*\pi*R^3=Volume~of~TWO~cones.\\ Hemisphere ~and ~cone~with ~same~radius~and~height~have~ the~ same ~surface~ area.\\ But ~Hemisphere~coast~is ~twice~for~Cone-cylinder.\\ So~Hemisphere~coast~is~=~TWO~cone~cost.\\ But~Cone-cylinder~cost~is~less~than ~Two~cone~cost, ~and~so~less~than~hemisphere~cost.

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A hemisphere and a cone with same radius and height don't have the same surface area.For hemisphere it is 2(pi)r^2 and for cone it will be (root2)(pi)r^2. @Niranjan Khanderia

Ankit Kumar Jain - 3 years ago

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Thank you. You are correct. I have changed the proof.

Niranjan Khanderia - 2 years, 12 months ago

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