Question 822055
Not enough data, and not reasonable data.
You must be missing some details of the problem.
A 1 meter wide cylindrical base means a monstrously large fountain. It would be appropriate for a city square, but not for the garden of any citizen.
A 1 meter diameter would be appropriate for the top cone of the fountain in someone's garden,
but even assuming 1 meter diameter and 1 meter base height, there are other design issues that are not specified.
Is the top cone part a solid cone?
If there is a cavity in the cone, is the wall of the cone uniformly thick? How thick?
How wide is the base?
Is the base a hollow stem? Or is it a solid cylinder made of concrete?
If it is hollow, how thick is the wall of the base cylinder?
Depending on design, it could be a complicated problem.
There is also the issue of how realistic you have to be.
Do you need to buy a whole number of liters of paint, or can you buy exactly the amount you need, even if it is a crazy decimal number, such as 0.17 liter?
What about the cement? Can you order 0.43 m^3?
 
The simplest, if ugly, design would be a cylinder with a conical hollow at the top.
In cross section, it would look like this:
{{{drawing(300,300,-55,55,-10,110,
line(-50,0,50,0),line(-50,0,-50,100),line(50,0,50,100),
line(-50,100,0,75),line(50,100,0,75))}}}
 
VOLUME:
The volume of that is the volume of the cylinder minus the volume of the cone.
If the cylinder and cone were 1 m in diameter (0.5m radius), with a cylinder height of 1m, and a cone height (depth) of 0.25m, the volume, in m^3 would be
{{{pi*0.5^2*1-pi*0.5^2*1/3=0.25pi-0.25pi/3=0.5236}}}
At $100/m^3, the cement to make that would cost $52.36.
 
SURFACE AREAS:
The surface to be painted would be the lateral surface of the cylinder,
calculates as {{{perimeter*height=pi*diameter*height}}} ,
plus the lateral surface of the cone,
calculated as {{{perimeter*height/2=pi*diameter*height}}}
For the measurements assumed above, the area in m^2 would be
{{{pi*1*1+pi*1*0.25/2=pi+pi/3=4pi/3=4.19}}}
 
COST:
The cost for 1 liter of paint and 0.53m^3 of concrete would be
$17.50 + $53.00 = $70.50
 
The fountain I would design, in cross-section, would look like this:
{{{drawing(300,400,-60,60,-20,140,
rectangle(-10,0,-8,100),rectangle(10,0,8,100),
line(0,96,-50,121),line(0,96,50,121),
line(-50,121,-45,121),line(50,121,45,121),
line(0,98.5,-45,121),line(0,98.5,45,121),
arrow(-5,127,-50,127),arrow(5,127,50,127),
locate(-4,130,1m),locate(11,53,1m),
arrow(14,54,14,100),arrow(14,46,14,0),
arrow(0,-3,-10,-3),arrow(0,-3,10,-3),
locate(-8,-4,0.20m),locate(-8,19,.16cm),
arrow(0,20,-8,20),arrow(0,20,8,20),
arrow(-10,121,-44.5,121),arrow(10,121,44.5,121),
locate(-10,124,0.90m)
)}}} The height of the cone is half the diameter.
 
VOLUMES:
The volume of cement in the cone is the volume of a cone with a 0.5m radius and a 0.25m height, minus the volume of a cone with a 0.45m radius and a 0.225m height. In {{{m^3}}} , it is
{{{(pi*0.5^2*0.25-pi*0.45^2*0.225)/3=pi*(0.06250-0.04556)/3=pi*.01694/3=0.01774}}}
The volume of cement in the base is the volume of a cylinder with a 0.1m radius and a 1m height, minus the volume of a cone with a 0.08m radius and a 1m height. In {{{m^3}}} , it is
{{{pi*0.1^2*1-pi*0.08^2*1=pi(.01-.0064)=0.036pi=0.01131}}}
The total volume of concrete is
{{{0.01774m^3+0.01131m^3=0.02905m^3}}} .
At $100/m^3 that would cost $2.91.
 
SURFACE AREAS:
The outside cone has a radius of 0.5m and a height of 0.25m.
Its slant height, in m, is
{{{sqrt(0.5^2+0.25^2)=sqrt(0.25+0.0625)=sqrt(0.3125)=0.559}}}
The perimeter of the base, in m, is {{{pi}}} , because the diameter is 1m.
The outside surface area of the cone, in {{{m^2}}} , is
{{{pi*0.559/2=0.878}}}.
The surface area of the cone insides plus the rim is about the same.
The outside surface area of the cylindrical base, in {{{m^2}}} , is
{{{pi*0.2*1=0.628}}}
The total area to be painted, in {{{m^2}}} , is about
{{{0.878+0.878+0.628=2.38}}}
That would require less than 1 liter of paint.
 
COST:
Buying 1 liter of paint, plus 0.03m^3 concrete would cost
$17.50 + $3.00 = $20.50