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3 years ago

What are the dimensions of the lightest open-top right circular cylindrical can that will hold a volume of 2197 cm cubed?

Mathematics

1 answer:

Zarrin [17]3 years ago

4 0

Answer:

radius = height ≈ 8.88 cm

Step-by-step explanation:

If the weight is proportional to the surface area, then we need to find the minimum surface area for the given volume.

The volume of the can is given by ...

V = πr²h . . . . . . for radius r and height h

The surface area includes that of the bottom and the sides:

SA = πr² +2πrh

Solving the first equation for h, we can write the second equation as ...

SA = πr² +2πr(V/(πr²)) = πr² +2V/r

The minimum area for the volume will be found where the derivative is zero.

d(SA)/dr = 2πr -2V/r² = 0

πr³ = V . . . . . multiply by r²/2 and add V

Then the can radius is ...

r = ∛(V/π)

and the height is ...

h = V/(π(V/π)^(2/3)) = ∛(V/π) = r

For the given volume, the dimensions are ...

r = h = ∛(2197/π) ≈ 8.8762 cm

The lightest can with that volume has a radius and height of about 8.88 cm.

_____

<em>Comment on cylinder aspect ratio</em>

You may notice that the lateral area is equal to twice the bottom area for this cylinder. If the can had a top, the minimum weight would be such that the lateral area is twice the total of top and bottom area. In that case, the height is equal to the diameter, not the radius.

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2 years ago

A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

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Answer:

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Step-by-step explanation:

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In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

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2 years ago

What are the dimensions of the lightest​ open-top right circular cylindrical can that will hold a volume of 2197 cm cubed​?

What are the dimensions of the lightest open-top right circular cylindrical can that will hold a volume of 2197 cm cubed?