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Kazeer [188]
3 years ago
11

An ice cream cone is filled with vanilla and chocolate ice cream at a ratio of 2:1. If the diameter of the cone is 2 inches and

the height is 6 inches, what is the volume of vanilla ice cream in the cone? (round to nearest tenth)
Mathematics
2 answers:
Marina86 [1]3 years ago
4 0
<span>An ice cream cone is filled with vanilla and chocolate ice cream at a ratio of 2:1. If the diameter of the cone is 2 inches and the height is 6 inches, what is the volume of vanilla ice cream in the cone? (round to nearest tenth) 
----
Total volume = pi*r^2*h = pi*1^2*6 = 6pi in^2
---------------------
Equation:
2x + x = 6pi

3x = 6pi
x = 2pi (chocolate volume)
2x = 4pi (vanilla volume)
</span>
Sophie [7]3 years ago
3 0
Volume of cone=(1/3)hpir^2

d/2=r
d=2
d/2=2/2=1=r
h=6

V=(1/3)6pi1^2
V=2pi in^3 is volume

ratio of 2:1

2+1=3
2pi=3 units
divide both sides by 3
2/3pi=1 unit

vanila=2 units
times 2/3pi by 2
4/3pi
aprox pi=3.141592
 4.188
round
4.2 in^3 of vanilla
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A metal cylinder can with an open top and closed bottom is to have volume 4 cubic feet. Approximate the dimensions that require
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Answer:

r\approx 1.084\ feet

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Step-by-step explanation:

<u>Optimizing With Derivatives </u>

The procedure to optimize a function (find its maximum or minimum) consists in :

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We know a cylinder has a volume of 4 ft^3. The volume of a cylinder is given by

\displaystyle V=\pi r^2h

Equating it to 4

\displaystyle \pi r^2h=4

Let's solve for h

\displaystyle h=\frac{4}{\pi r^2}

A cylinder with an open-top has only one circle as the shape of the lid and has a lateral area computed as a rectangle of height h and base equal to the length of a circle. Thus, the total area of the material to make the cylinder is

\displaystyle A=\pi r^2+2\pi rh

Replacing the formula of h

\displaystyle A=\pi r^2+2\pi r \left (\frac{4}{\pi r^2}\right )

Simplifying

\displaystyle A=\pi r^2+\frac{8}{r}

We have the function of the area in terms of one variable. Now we compute the first derivative and equal it to zero

\displaystyle A'=2\pi r-\frac{8}{r^2}=0

Rearranging

\displaystyle 2\pi r=\frac{8}{r^2}

Solving for r

\displaystyle r^3=\frac{4}{\pi }

\displaystyle r=\sqrt[3]{\frac{4}{\pi }}\approx 1.084\ feet

Computing h

\displaystyle h=\frac{4}{\pi \ r^2}\approx 1.084\ feet

We can see the height and the radius are of the same size. We check if the critical point is a maximum or a minimum by computing the second derivative

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We can see it will be always positive regardless of the value of r (assumed positive too), so the critical point is a minimum.

The minimum area is

\displaystyle A=\pi(1.084)^2+\frac{8}{1.084}

\boxed{ A=11.07\ ft^2}

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