a company is manufacturer of cylinrical aluminium tins.the manager plans 2 reduce d cost.?

Question

the cost is proportional 2 d area of d aluminium sheet.d volume tat each tin can hold is 1000cm3 (1liter)....tat my project work...waiting a genius 2 help me!!!
1)determine d value of h,r and hence calculate d ratio of h/r when d total surface area of each tin is minimum.here,hcm denotes d height n rcm d radius of d tin.

Answer 1

Volume = (pi)(r^2)h = 1000 so h = 1000/((pi)r^2)

Total surface area,
a = 2(pi)rh + 2(pi)r^2

substitution of h yields:
a = (2000/r) + 2(pi)r^2

To find the minimum value of r we must set the first derivative of a with respect to r, da/dr = 0

da/dr = (-2000/r^2) + 4(pi)r

For da/dr = 0
4(pi)r = 2000/r^2
(pi)r^3 = 500
so r = cbrt(500/(pi)) where cbrt = cube root
r = 5.419 cm (3 decimal places)

To check this is a minimum value we find the second derivative od a with respect to r, d^a/dr^2:

d^a/dr^2 = (4000/r^3) + 4(pi)
substitution for r^3 = 500/(pi) yields
d^a/dr^2 = 12(pi) This is a positive number proving that r is a minima.

We can now go back to one of our earlier equations
h = 1000/((pi)r^2) and substitute r:
h = 10.840 cm (3 decimal places)

To find the ratio of h/r it is best to use the equations to get an exact answer

h/r = 1000/((pi)r^3) but r^3 = 500/(pi)
so h/r = 2

Hope that helps!

Answer 2

If I understand correctly you want a cylinder that can hold 1000cm3, but has the smallest surface.
Every Celinda can hold r^2*Pi*H cm3, where r is half of the radius of the base, and H is its height.
so r^2*H=1000/Pi
so H=1000/(Pi*r^2)
On the other hand, the surface is 2*Pi*r(H+r)
So you are looking for an r so that for a function 2*Pi*r(1000/(Pi*r^2)+r)=y y is minimum (y represents the surface of the tin)
2000/r+2*Pi*r^2=y
I think this is called derivation
4*Pi*r-2000/r2=y'
Looking for critical points
4*Pi*r-2000/r2=0
r^3*Pi*4-2000=0
r^3*Pi=500
r=5,41926... is one, and r=0 is the other
For r<0 y' is negative, so y is decreasing
for r>0 and r<5,41926 y' is still negative so r isn't an actual critical point.
for r>5,41926 y' is positive so y is rising.
so minimum that y can be is for r=5,41926... cm, or r is cube root of 500/Pi, and that is the best r.
H is 10.838524....cm

Answer 3

The total area S of the cylinder is S = 2pi*r^2 + 2pi*r*h. The first parcel is the area of the 2 circular bases and the second is the lateral area of the cylinder. The cylinder volume is fixed in 1000 cm3. We have V = pi*r^2 *h, so that h = V/(pi*r^2). Plugging this in the expression of S, we get S = 2pi*r^2 + 2pi*r*V/(pi*r^2) = 2pi*r^2 + 2V/r. Now, we have to find r so that S is minimum. We have dS/dr = 4pi*r - 2V/(r^2), so that dS/dr = 0 => 4pi*r^3 = 2V => r = r* = (V/(2pi)^(1/3), where r* is the desired value of r. We see this is a minimum point, because dS/dr <0 for 0 < r < r* and dS/dr >0 for r > r* . Substituting in the expression of h, we get h* = V//(pi*r^2+ = V /pi * (2pi//V)^(2/3)= (4V/pi)^(1/3). Therefore, h/d = 8^(1/3) = 2, and this ratio is independent of the volume V.