Question

A cylindrical container, with the top end opened, is to be constructed to hold a volume of
1.50 m³. Find the dimensions of the container such that the material used would be a minimum. Give your answers to 2 decimal places.

Answer 1

i think here is no one except us:)

Answer 2

What are you trying to optimize. In your mind's eye, visualize what graph you're trying to find the minimum of. In this case, you want the least amount of material to hold 1.50 m^3 right? So then you will want the least surface area, right? So when the derivative of the graph of surface area is zero, you'll know that's the point where the least amount of material is used!

Now all you need to do is make a formula for area and see that it's in terms of radius and volume. Luckily, you can make the equation all in terms of one variable through the information given and the formula for volume. That way when you take the derivative of area, you can do it with respect to one variable, the radius or the height.

Answer 3

That second part I meant to say height at the end of the first sentence, not volume.:

"Now all you need to do is make a formula for area and see that it's in terms of radius and height. "

Answer 4

.78 is a number significant to this problem, and that's for you to figure out how.

Answer 5

\[V=1.5=\pi ^{2} rh\]
\[r ^{2}h=\frac{1.5}{\pi}\]
\[rh=\frac{1.5}{\pi r}\]
\[SA=\pi r ^{2}+2\pi rh\]
Substituting for rh gives
\[SA=\pi r ^{2}+\frac{2\pi \times 1.5}{\pi r}=\pi r ^{2}+\frac{3}{r}\]
Now we have an equation for surface area in one variable (r). The next step is to differentiate SA with respect to r and solve the equation f'(SA) = 0. When a value for r is obtained substitute the value in f''(SA). If the result is positive then this value for r gives a minimum amount of material for the cylindrical container.

Answer 6

He said the same thing as me, except he's holding your hand through the whole fun part of the problem that he won't be able to help you with constructing and imagining on the test.

Answer 7

i had done these steps
V=1.5=π2rh

r2h=1.5π

rh=1.5πr

and kinda lost Substituting in

Answer 8

@mathnoob1 The first equation in my previous posting should read
\[V=1.5=\pi r ^{2}h\]

Answer 9

dSA/ dr = 2πr - 3/r^2 ?

Answer 10

@mathnoob1 Your differentiation is correct :). Now put the result equal to zero and solve to find r.

Answer 11

2πr-(3/(πr²))=0
2πr=(3/(πr²))
2πr³=(3/π)

Answer 12

@mathnoob1 you added an extra pi. Start again with the result of your differentiation and you will get
\[2\pi r=\frac{3}{r ^{2}}\]

Answer 13

argh it [3]√((3/(2π))) !!

Answer 14

thank you Kainui and kropot72 for your time and patience =)

Answer 15

You're welcome :)