A solid is formed by adjoining two hemispheres to ends of a right circular cylinder.?

Question

The total volume of the solid is 12 cubic centimeters. Find the radius of the cylinder that produces the minimum surface area.

Answer 1

Use r for the radius and h for the height of the cylinder. Write an expression for the total volume in terms of r and h and equate it to 12. Rearrange this to get h in terms of r. Now write down the total surface area in terms of h and r and substitute for h what you found before. You now have total area A as a function of r. Differentiate this and continue in the usual way.

Answer 2

r = [9 / pi]^(1/3) cm = 1.420248085... cm.

Let the radius of the cylinder be ' r ' and its length ' h.' Then its volume ' V ' satisfies :

V = pi r^2 h + (4 pi / 3) r^3 = 12, so that h = [12 - (4 pi / 3) r^3] / [pi r^2] ....(A).

Its surface area ' S ' is given by:

S = 2 pi r h + 4 pi r^2, so, by (A),

S = {2 pi r / [pi r^2]} [12 - (4 pi / 3) r^3] + 4 pi r^2 = 24 / r + (4 pi / 3) r^2 ....(B).

dS / dr = - 24 / r^2 + 8 pi r / 3 = 0 when - 3 / r^2 + pi r / 3 = 0, i.e.
r^3 = 9 /pi.

Thus r = [9 / pi]^(1/3) cm = 1.420248085... cm gives a stationary value for the surface area.

d^2 S / dr^2 is POSITIVE (always), because it consists of two positive terms. So yes, it is at a MINUMUM at this value of r.

[ CHECK? : We can check that at least this last part was done correctly by evaluating S at r_stat = 1.420248085, and at stat +/- 0.01. (Such a check showed that there had been a numerical error in evaluating the correct functional form, in my original posting.)

For r_stat, S_min = 25.34768424... .
For r_stat + 0.01, S = 25.34893502... .
For r_stat - 0.01, S = 25.34894682... .

These values confirm the claim of a "minimum" at r_stat. ]

Live long and prosper.

Answer 3

See the quantity of the tank has been fastened.this could supply a relation between the two variable parameters it quite is the peak and radius.The equation would be 5500=4pi(r^3)/3+pi(r^2)h h=[5500-4pi(r^3)/3]pi*r^2 now the 2d realtion will come from the fee and are a relation fee=2(4pir^2)+2(pi)rh =2(4pir^2)+2(pi)r*=[5500-4pi(r^3)/3]pi... it quite is a function between radius and fee now diffrentiate the function and equate it to 0.this supply you with the fee of r for minimum fee. making use of first relation you are able to locate h via substituting the fee of r.