calculus! please help! complicated i dont get it?

Question

For the next problems you will need to come up with an equation for the quantity you are trying to minimize in terms of a single independent variable. then i would like you to graph the function on your calculator and use the "calculate minimum" capability of the calculator to find the answer.

1)A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. The total volume of the solid is 12 cubic centimeters. find radius of cylinder that produces the minimum surface area.
2) An industrial tank of the same shape in (1) must have a volume of 3000 cubic feet. The hemispherical ends cost twice as much per square foot of surface area as the sides. Find the dimensions that will minimize cost.

my teacher gave me that can please some one help because i don't know where to start! Thanx!

please anyone help!
i realy need this thanx

for 1) i need radius
i cant solve for h

Answer 1

1) The total volume :

V = 12 = (4/3)πr^3 + πr²h, where h is the length of the cylinder and r its radius.
(4/3)πr^3 is the volume of the two hemispheres and πr²h the volume of the cylinder part.

the total surface, by the same reasoning is:
S(r) = 4πr² + 2πrh

from the first equation we get πrh = 1/r*(V - (4/3)πr^3)
πrh = 12/r - (4/3)πr²
substitute in the surface calculation:
S(r) = 4πr² + 2(12/r - (4/3)πr²)
S(r) = 4/3πr² + 24/r
differentiate S(r) :
dS(r)/dr = 8/3πr -24/r²
dS(r)/dr = 0 when (8/3)πr = 24/r², or r^3 = 9/π
your minimum is reached when r = (9/π)^(1/3)
r = 1.46 cm

2) Instead of minimizing the surface, you are minimizing the cost.
V = 3000 = (4/3)πr^3 + πr²h
C(r) = 2S1(r) + S2(r) where S1 is the surface of the hemispheres and S2 the surface of the cylindrical part.
C(r) = 8πr² + 2πrh
πrh = 3000/r - (4/3)πr²
C(r) = 8πr² + 2(3000/r - (4/3)πr²)
C(r) = (16/3)πr² + 6000/r
dC(r)/dr = (32/3)πr - 6000/r²
dC(r)/dr = 0 when (32/3)πr = 6000/r², or r^3 = (1125/(2π))
r = (1125/2π)^(1/3) = 5.64 feet.
Replace in the volume equation :
3000 = (4/3)πr^3 + πr²h = (4/3)π*1125/(2π) + π*(1125/2π)^(2/3)h
3000 = 750 + π*(1125/2π)^(2/3)h
h = 2250/( π*(1125/2π)^(2/3)) = 22.5 feet.

Answer 2

1) Volume total = Vol cyl + vol of 2 hemispheres

2 hemispheres = 1 sphere

V = pi*r ^ 2 *h + (4/3) * pi* r ^ 3

SA = 2*pi*r*h + 4*pi*r ^ 2........[ SA of cyl, minus the ends, + SA of a sphere ]

to min SA, usually need to take the derivative, but you can also graph it, and use the calculator to find Max or min......but first eliminate either r, or h using volume formula...it is easier to eliminate the "h" here :

since 12 = pi*r ^ 2 *h + (4/3) * pi* r ^ 3 , solve for h:

h = [ 12 - (4/3) * pi* r ^ 3 ] / pi*r ^ 2

SA = 2*pi*r* [ 12 - (4/3) * pi* r ^ 3 ] / pi*r ^ 2 ]] + 4*pi*r ^ 2
SA = 24/ r - (8/3)*pi* r^2 + 4*pi*r ^ 2
graph it......I used a window of x: [ 0,3], and Y: [ 20,40]

I got x [ which = r ] as 1.420 [ corrected from the above 1.46 ] and y = SA = 25.348
see what you get..............[[[ this is the same as the above ( 9/pi)^(1/3).........


2) V = 3000 = above eqn

cost = k*{SA cyl.] + 2k*{SA of sphere]
note: k = cost per sq. ft of the side..it will disappear in the derivative, or set = to 1 if graphing.........
C = k * 2*pi*r*h + 2k*4*pi*r ^ 2.= 2k*pi*r*h + 8k*pi*r ^ 2.
by volume........3000 = pi*r ^ 2 *h + (4/3) * pi* r ^ 3

so h = [ 3000 - (4/3) * pi* r ^ 3 ] / pi*r ^ 2

and cost: C = [ 6000k/r) - (8/3)*pi*k*r^2 + 8k*pi*r^2

I got x = radius = 5.64 at y = cost = 1596.8
as a minimum......h = 27.4
checking.the volume worked out high, at 3489.65, but I gotta go do some work, so I don't know if it was an error on my work, or a rounding error.........will check again later.............

I'm back........I got h = 22.5 when I re-entered it on my Calculator, and V = 2999.98, using my r and h values....must of hit the wrong button on the calculator.......that's what happens when you are in a hurry............. :-)

note: on the derivative on the 2nd answer given below mine, there is a mistake on simplifying his work.......he should get 18,000 m - 8pi*m*r^2 in the ( ), but it still divided by 3r, and not just r.........
here it is corrected:

C = 8m*pi*r^2 + 2m*pi*r*( 3000 - (4/3)*pi*r^3 ) / [ pi*r^2]
simplifying, you get C = 8m*pi*r^2 + 6000m/r - (8/3)m*pi*r^2
or C = (16/3)m*pi*r^2 + 6000m/r

dC/dr = 32m*pi*r / 3 - 6000m / (r^2).....graphing the derivative, you would like C ' = 0, which takes place at r = 5.636, or 5.64, as I gave earlier.......

Answer 3

The area of the caps is Ac = 4*pi*r^2
The area of the cylinder is A = 2*pi*r*h where h = height of cylinder

Total area = At = 2pi*(2r^2+r*h)

Take first derivative & set to zero

d At/dr = 0 = 4r+h ---> r = h/4 note this is the min. since the second derivative is > 0.

Now using a fixed volume we can find h. Volume is

V = 2*(2/3*pi*r^3) +pi*r^2*h
= 4/3*pi*h^3/64 + pi *h^3/16
= pi*h^3*16/192 = 1/12*pi*h^3

Now V=12cm^3 so 12 = 1/12*pi*h^3 ---> h=(144/pi)^(1/3) = 3.58 cm

r = h/4 = 0.9 cm

For next problem, you want to miniminze cost. The cost function is:

C = 2*Ac+A = 2*4*pi*r^2+ 2*pi*r*h

dC/dr = 0 =16*pi*r +2*pi*h ---> r = h/8

Now repeat above calculation for volume, solve for h, and you get r and you're done.

Answer 4

V = 4/3 π r³ +π r² h= 12 then h = (12- 4/3 π r³ )/π r²
A = 4π r² + 2πr h =4π r² +2πr(12- 4/3 π r³ )/π r²
A= 4π r² +(24- 8/3 π r³ )/r = (4π r³ +24 - 8/3 π r³ )/r
A= (12π r³ +72 - 8 π r³ )/3r
A= (4π r³ +72 )/3r
dA/dr = [3r(12πr²)-(4π r³ +72 )3]/9r² =0
=12π r³-4π r³ -72
then 8π r³ =72 then π r³ =9
r³= 9/ π
then r = 1.42025 cm

2)costC= 8mπ r²+m2πr(3000 - 4/3 π r³ )/π r²
C= [8mπ r³+m(6000-8/3 π r³)]/r=[24mπ r³+m(18000-8 π r³)]/r
C= [16mπ r³ +18000m]/r
dC/dr =[m(48π r³- (16π r³ +18000)]/r² =0
then 32r³=18000 then r=8.255 ft
h=(3000- 4/3 π r³ )/π r² =3 ft

Answer 5

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