Question

Find the dementions (radius and height) of a cylindrical can in order to minimize the surface area and have a fixed volume of 1000 cubic centimeters
help please i dont get it

Answer 1

do you have any idea how to do this?

Answer 2

yes, do you know the Method of La Grange?

Answer 3

No i dont, my math teacher never mentioned that.

Answer 4

Ok. l'll show you by using it.

Answer 5

okay thank you so much

Answer 6

you still there?

Answer 7

?

Answer 8

A=surface area = 2*pi*r^2 + 2pi*r*h, including ends of can and sides
Constraint: V = volume = pi*r^2*h = 1000 cm^2

dA/r = 4*pi*r + 2*pi*h= L * dV/r = L * 2pi*r*h (eq 1)
dA/h = 2*pi*r = L * 2*r^2 (eq 2)

Need to first find h and r in terms of L then plug into constraint to find minimized surface area.

from eq 1:
2*pi*r = L * 2 * r^2
2*pi/(2L) = r

from eq 2:
4*pi*r + 2*pi*h = L * 2pi*r*h
4*pi*r = L * 2pi*r*h - 2pi*h
= h(L*2*pi*r - 2*pi)
h = 4*pi*r/(L*2pi*r - 2*pi)
= 4*pi*r/( 2*pi (L*r - 1) )
= 2*r/(L*r - 1)
= 2*(2*pi/(2L))/(L*r - 1)
= 8*pi*L/(L*(2*pi/(2L)) - 1)
= 8*pi*L/(pi - 1)

Using constraint:

pi*r^2*h = 1000
pi*(pi/L)^2*8*pi*L/(pi - 1) = 1000
8Lpi^3/(pi - 1) = 1000

L = (pi - 1)*1000/(8*pi^3)
= (pi - 1)*250/(4*pi^3)
= 4.32 (approx)

Plug back into r and h:

2*pi/(2L) = r
r = pi/L = pi/4.32 = 0.73

h = 8*pi*L/(pi - 1)
= 50.70

Plug into surface area (A) to find minimized value:

A = 2*pi*r^2 + 2pi*r*h
= 2*pi*0.73^2 + 2pi*0.73*50.70
= 236.00

Ans. 236.00 cm^2

Please check my arithmetic, but the essence of The La Grange is all here. Easy process but lots of calculations.

Answer 9

but isnt it suppose to be cm^3 not cm ^2?

Answer 10

Area is square cm and volume is cubic, because area is 2 dim and volume is 3 dim. Also, question asks for r and h not area. I also have an error in the first part, I'll correct below.

Answer 11

so can you list what the radius and height will be? i dont know where the 236.00 belongs hehe

Answer 12

A=surface area = 2*pi*r^2 + 2pi*r*h, including ends of can and sides
Constraint: V = volume = pi*r^2*h = 1000 cm^2

dA/r = 4*pi*r + 2*pi*h= L * dV/r = L * 2pi*r*h (eq 1)
dA/h = 2*pi*r = L * dV/h = L * pi*r^2 (eq 2)

Need to first find h and r in terms of L then plug into constraint to find minimized surface area.

from eq 2:
2*pi*r = L * pi * r^2
2/L = r

from eq 1:
4*pi*r + 2*pi*h = L * 2pi*r*h
4*pi*r = L * 2pi*r*h - 2pi*h
= h(L*2*pi*r - 2*pi)
h = 4*pi*r/(L*2pi*r - 2*pi)
= 4*pi*r/( 2*pi (L*r - 1) )
= 2*r/(L*r - 1)
= 2*(2/L))/(L*r - 1)
= 4/(L*(2/L)) - 1)
= 4/(2 - 1/L)

Using constraint:

pi*r^2*h = 1000
pi*(2/L)^2*4/(2 - 1/L) = 1000
16*pi/(2L - 1) = 1000
16*pi/1000 = 2L - 1
2L = 4pi/250 + 1
L = 2pi/250 + 1
L = 1.03 (approx)

Plug back into r and h:

2/L = r
r = 2/1.03 = 1.95 cm

h = 4/(2 - 1/L)
= 3.87 cm

Plug into surface area (A) to find minimized value:

A = 2*pi*r^2 + 2pi*r*h
= 2*pi*1.95^2 + 2pi*1.95*3.87
= 71.31 cm^2

Ans. r = 1.95 cm, h = 3.87 cm

Please check my arithmetic, but the essence of The La Grange is all here. Easy process but lots of calculations.

Answer 13

To see if area is minimum, try another value of r and solve for h in constraint.
Area should be larger.

Answer 14

is there any way you can help me with the other question i asked? i promise its not this much work hehe.