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Find the maximum volume of a cylindrical soda can such that the sum of its height and its circumference is 120 cm.

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LammettHash
The volume of such a can with base radius [tex]r[/tex] and height [tex]h[/tex] would be

[tex]V(r,h)=\pi r^2h[/tex]


We desire the base's circumference and the can's height to add to 120, i.e.

[tex]2\pi r+h=120\implies h=120-2\pi r[/tex]

Substituting this into [tex]V(r,h)[/tex] allows us to reduce the volume to a function of a single variable [tex]r[/tex]:

[tex]V(r,120-2\pi r)=V(r)=\pi r^2(120-2\pi r)=120\pi r^2-2\pi^2r^3[/tex]

Taking the derivative with respect to [tex]r[/tex] yields

[tex]V'(r)=240\pi r-6\pi^2r^2[/tex]

Set this equal to 0 and find any critical points:

[tex]240\pi r-6\pi^2r^2=0\implies 6\pi r(40-\pi r)=0\implies r=0\text{ or }r=\dfrac{40}\pi[/tex]

This suggests the can will have maximum volume when its radius is [tex]\dfrac{40}\pi\approx12.7[/tex] cm, which would give a volume of about 20,371 sq. cm.

The maximum volume of cylinder is 20371.83 cm³.

Let us consider that, radius of cylinder is r and height is h .

Volume of cylinder is,

                [tex]V=\pi r^{2}h[/tex]

Since, the sum of its height and its circumference is 120 cm.

So,    [tex]h+2\pi r=120[/tex]

         [tex]h=120-2\pi r[/tex]

Substituting value of h in volume equation of cylinder.

         [tex]V=\pi r^{2}(120-2\pi r)\\\\V=120\pi r^{2}-2\pi^{2}r^{3}[/tex]

For maximum volume, differentiate above expression with respect to r and equate with zero.

          [tex]\frac{dV}{dr} =240\pi r-6\pi^{2}r^{2} =0\\\\6\pi r(\pi r-40)=0\\\\r=0, r=\frac{40}{\pi}[/tex]

So, radius of cylinder is,  [tex]r=\frac{40}{\pi}[/tex]

Now,  [tex]h=120-2\pi (\frac{40}{\pi} )=80cm[/tex]

Substituting value of r and h in volume formula of cylinder.

    [tex]V=\pi (\frac{40}{\pi} )^{2} *40=20371.83cm^{3}[/tex]

Value of [tex]\pi=22/7=3.14[/tex]

Thus, the maximum volume of cylinder is 20371.83 cm³.

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