how to solve this problem?

Question

the Muhibbah company is a manufacturer of cylindirical aluminium tins. The manager plans to reduce the cost of production. the production cost is propotional to the area of the aluminium sheet used. the volume that each tin can hold is 1000 cubic cm
(1 liter).
1. determine the value of h, r and hence calculate the ratio of h over rwhe the total surfacearea of each tin is minimum. here, h cm is denotes the height and r cm the radius of the tin.
2. The top and bottom pieces of the tin of height h cm re cut from square shaped aluminium sheets.Determine the value for r,h and hence calculate the ratio h over r so that the total area of the aluminium sheet used for making the tin is minimum
3.Investigate cases where the top and bottom surfaces are cut from
(a)equilateral triangle
(b)regular haxagon
find the ratio of h over r for each case

Answer 1

V = 1000 = pi r^2 h, then you have the condition that

h = 1000 / (pi r^2)

This condition will be used in all cases. The surface used in each case is

1. A(r,h) = 2 pi r h + pi r^2
2. A(r,h) = 2 pi r h + 8 r^2
3a. A(r,h) = 2 pi r h + 6 3^(1/2) r^2
3b. A(r,h) = 2 pi r h + 12 3^(-1/2) r^2

just put the above condition to eliminate the dependence on h

1. A(r) = 2000/r + pi r^2
2. A(r) = 2000/r + 8 r^2
3a. A(r) = 2000/r + 6 3^(1/2) r^2
3b. A(r) = 2000/r + 12 3^(-1/2) r^2

next thing to do is calculate the first derivative, equals to 0 and solve for r. You have to do the same derivative en each case, the only different is the constant in the secon term, so you can do a general procedure to solve all at once. Say "a" a constant value a = pi in the first case and so on

A(r) = 2000/r + a r^2
dA/dr = -2000/r^2 + 2 a r = 0
r^3 = 1000/a
r = 10/a^(1/3)
h = 1000 / (pi r^2) = 10 a^(2/3) / pi
h/r = a / pi


1. a = pi then r = 10/pi^(1/3), h = 10/pi^(1/3) & h/r = 1
2. a = 8 then r = 5, h = 40/pi & h/r = 8/pi

the same for the others with "a" value

3a. a = 6 3^(1/2)
3b. a = 12 3^(-1/2)

for simplicity in the last ones, take your calculator and obtain de value of "a" and then put it in the eqs for r, h and h/r above.

Let me know if you have any problem

Answer 2

A more realistic version of this problem would have been to have a different cost associated with the two distinct parts of the container. The top and bottom of a can probably have a different cost per square inch of material than the side of the can. That said the method to solve problem 1 as stated is to write two formulas involving the R=radius, H=height, A=area of aluminum of the can.
The first one I would use would be the volume of the can.
Equation #1 1000 = Pi * R * R * H
The second would be a formula for the Area of aluminum of the can.
Equation #2 A= 2 * Pi * R * R + 2 * Pi * R * H

Solve Equation #1 for H so that it can be substituted into Equation #2 to result in an equation that has only A and R.

Equation #1S H=1000 / Pi / R / R

Equation #2S A= 2 * Pi * R * R + 2000 / R

Take the derivative of Equation #2S and determine what value of R results in dA/dR =0 When R is a positive real number.

Substitute that value of R into Equation #1S to determine what H is.

R should be a little more than 5.3
H should be a little more than 11.1
H over R should be a little more than 2

That is how to solve problem 1.

Problem 2 would be solved by a similar process only the Equation #2 would be slightly different

Equation #2 A= 8 * R * R + 2 * Pi * R * H
This version of the general problem does not completely describe the real world costs either as it ignores the difference in cost of cutting a round piece from cutting a straight line.

Problem #3 would again be solved with slightly different Equation #2's. Sustitute the area for each of the shapes (triangle and hexagon) for the portion of Equation #2 representing the top and bottom of the can. Problem #3 seems to imply that the sheets of metal which we are starting, are sufficiently large that we can ignore the scrap left over from cutting the triangles or Hexagons(not haxagon)