What are the dimensions of a cylindrical jar with surface area 1200 squared centimetres and the greatest volume?

b ) Sketch the jar and label its dimensions.

c) What is the maximum volume of the cylinder?

This post was edited by RollOneUp on Dec 9 2022 12:02pm

please

Are you in calculus or in algebra?

There’s a solution using either I just need to know what class.

For convenience, lets chose conventional symbols for the dimensions of the jar: Let r be the radius of the bottom (which is a circle), let h be the height.
Then the volume is given by V = h*pi*r^2, the total surface is given by S = 2*pi*r*h + 2*pi*r^2 (assuming that the jar is closed at the top and the bottom, that is not formulated well in the task...).

We can rearrange the second formula to solve for h:
h = (S - 2*pi*r^2)/(2*pi*r)
Inserting this into the first formula yields the volume in dependency of just one variable, r:
V = (S - 2*pi*r^2)/(2*pi*r)*pi*r^2 = (S - 2*pi*r^2)*r/2 = (S*r - 2*pi*r^3)/2

To get the maximum volume, we take the first derivative of V with respect to r and set it to be 0:
dV/dr = (S - 6*pi*r^2)/2 = 0
Solving for r yields:
r = +/- sqrt( S/(6*pi) ) =: r_0
The negative solution can be ignored since r has to be non-negative.

Since we are dealing with a function of a restricted domain, we have to test all local extrema (zeros of the first derivative) as well as the borders of domain:
V(r=0) = 0 (i.e. not a maximum value of V)
lim[r -> infinity] V(r) = - infinity (i.e. not a maximum value of V)
V(r_0) = S^(3/2)/( 3*sqrt(6*pi) ) (i.e. this is indeed the maximum value of V)

The maximum volume can thus be calculated just by plugging in the given value of S:
V(r_0) = 3192 cm^3 = 3.192 L


hoping i did no typos in any of the formulas...

The algebra solution (no calculus)

You have to realize that a square is the optimal rectangle. So

you need to sketch a cylinder but the idea is that it is inside a square (draw a cylinder with a square around it). When you do this you can hopefully see that the height of your cylinder is equal to the diameter or twice the radius.

You can label the sides as h and 2r and note that h=2r

Will nice you make this deduction you can actually do part c.

C)


SA = 2 pi r^2 + 2 pi r h

Plug in your given 1200 constant for SA

1200 = 2 pi r^2 + 2 pi r h

We said from part b that h = 2r

1200 = 2 pi r^2 + 2 pi r 2r

1200 = 2 pi r^2 + 4 pi r^2

1200 = 6 pi r^2

Divide by 6

200 = pi r^2

200/pi = r^2

Now square root

Sqrt(200/pi) = r

r = 7.97884560803

We said before that h=2r

h = 2r = 2(7.97884560803) = 15.9576912161

Now that we have h and r we can find volume.

V = pi r^2 h

V = pi (7.97884560803)^2 (15.9576912161)

V = 3191.53824322 cm^3

i can use this answer for A also ? which is the dimensions since you determined the H and R during this

I appreciate all of your help

The second part I am also struggling with

this is the question

2. Rhonda has been hired by a local construction company to design a container that would hold 50ft3 of sand. The container will permanently remain on the company’s lot but should be designed so that the sand is easily accessible. Rhonda was not given specific instructions about the type or shape of container, only a few general guidelines:

A )Create a design for at least two different types of containers.
B ) Determine the dimensions of the containers such that the amount of material used to create the container is kept to a minimum.
C ) Include a realistic diagram of the containers that includes the dimensions.
D )Determine the cost to produce each container given that the cost of the materials is $0.40 per square foot.
E ) Make a recommendation for one (1) container that you feel is the best choice and explain the reasons for your choice. Note: Although cost is to be kept at a minimum, the cheapest container to produce may or may not be practical to use. Your explanation should provide details as to why you are recommending one container versus another.

What I would need help with here are B and D
I would probably choose a cylinder shape of container and probably a rectangular prism for the second container

The most optional prism is a cube

The most optimal cylinder is one with h = 2r

I believe the most optimal shape would be a sphere, but it is too impractical to consider. Optimizing solely for the purpose of minimizing material is not practical. 1 cubic foot is sand weighs around 101lbs. Imagine 50 cubic feet of sand in a spherical container. It would never sit still and if it was ever moved down an incline then it could cause serious damage. —imagine parking on a steep hill without your emergency break on.

Cylinders and prisms are used so often because they are practical. Easy to stack easy to move easy to tesselate a space with minimal waste.

Volume cube : s^3

50 = s^3

50^(1/3) = s

3.68403149864 = s

Surface area cube = 6s^2

Surface area = 6(50^(1/3))^2 = 81.4325284978 sq ft

Volume cylinder = pi r^2 h

Note that h= 2r


Volume cylinder = pi r^2 (2r)

Volume cylinder = 2 pi r^3

50 = 2pi r^3

(50/2pi) = r^3

(50/2pi)^(1/3) = r

1.99647271233 = r

Surface area cylinder = 2 pi r^2 + 2pi r h

Surface area cylinder = 2 pi r^2 + 2pi r (2r)

Surface area cylinder = 6 pi r^2

Surface area cylinder = 6 pi (1.99647271233)^2

Surface area cylinder = 75.1325069828

Now that we have surface area we can just multiple by the cost of the material

Cost per cube = 81.4325284978 (0.4) = 32.5730113991

Cost per cylinder = 75.1325069828 (0.4) = 30.0530027931