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4 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]4 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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3 years ago

What is the period and midline?

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$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\

\end{array}\qquad$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\
\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\


\end{array}$

$\bf \begin{array}{llll}
\bullet \textit{function period}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}
$

so if you notice yours $\bf \begin{array}{llll}
3.2cos&\left( \frac{5}{3}\theta \right)+&6.1\\
&\ \uparrow&\uparrow \\
&B&D 
\end{array}$

now.. normally the function $\bf 3.2cos&\left( \frac{5}{3}\theta \right)$
has a D value of 0, or no vertical shift, and the amplitude and the period simply make the wave taller and thinner, but the midline is still the x-axis

now, with D = 6.1, that moves the midline up vertically that much

now.. the period, well, B = 5/3, normal period of cosine is $2\pi$
so, the new period will be $\bf \cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{5}{3}}\implies \cfrac{6\pi }{5}$

notice the picture below
the vertical shift by the D component, or 6.1, moved the midline to y = 6.1 :)

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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​?

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