AP calculus! please help! complicated?

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 volume of the solid = volume of cylinder + volume of two hemispheres. So, you have V = pi * r^2 * h (the length of the cylinder) + (4/3) * pi * r^3. This condenses to:
V = pi * [(h * r^2) + (4/3) * r^3)] = 12.

This equation cannot be put into terms of only one independent variable, because the length of the cylinder (h) and the radius (r) are independent of one another. One of the two variables needs to have a set value first.

2) Once you find the length of the cylinder and the radius, then calculating the surface area is simple. For the cylinder, S = 2 * pi * r * h; for the spheres S = 4 * pi * r^2.

Answer 2

Good problem but you need to provide info about the length of the cylinder other wise I can't give you any exact answers.

in general you just need to take the first derivative of the function for the surface area and find the value of r with will result in the first derivate evaluating to zero.

Answer 3

1) As radius decreases, surface does too. So:
radius = smallest number > 0

2) As radius decreases, surface does too. So:
radius = smallest number > 0 will minimize the cost.

Answer 4

you will basically truthfully want the TI-89 in case you have been making plans on majoring in math, engineering, or a math-heavy technological know-how like physics in college. via fact it isn't the case, i might say the TI-80 4 Plus. in case you basically like it for a three hundred and sixty 5 days, the TI-80 3 Plus will do as properly.

Answer 5

1.
12=4pi*r^3/3+pi*r^2h
=pi*r^2(4r/3+h)

h=12/pi*r^2 - 4r/3

A=4pi*r^2+2pi*rh
=2pi*r(2r+h)
=2pi*r(12/pi*r^2 +2r/3)
=24/r+4pi*r^2/3

dA/dr=-12/r^2+8pi*r/3=0
r^3=36/8pi
r=0.5cbrt(36/pi)

2.
cost,C=4pi*r^2+4pi*rh
=4pi*r(r+h)
=4pi*r(3000/pi*r^2 - r/3)

dC/dr=0, and u got ur r already