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Elanso [62]
3 years ago
14

Find the surface area

Mathematics
1 answer:
mixas84 [53]3 years ago
7 0
Surface Area formula of a cylinder:
S.A = 2 × Area of base + Perimeter of base × Height of prism
       = 2\pir² + 2\pirh
       = 2\pi(5)² + 2\pi(5)(18)
       = 50\pi + 180\pi
       = 230\pi in² ( in terms of \pi )
       = 722.566 in²
       = 722.7 in² ( to the nearest tenth )
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6 0
3 years ago
Hello people ~
Luden [163]

Cone details:

  • height: h cm
  • radius: r cm

Sphere details:

  • radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

<u>Using Pythagoras Theorem</u>

(a)

TO² + TU² = OU²

(h-10)² + r² = 10²                                   [insert values]

r² = 10² - (h-10)²                                     [change sides]

r² = 100 - (h² -20h + 100)                       [expand]

r² = 100 - h² + 20h -100                        [simplify]

r² = 20h - h²                                          [shown]

r = √20h - h²                                       ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (\sqrt{20h - h^2})^2  \  ( h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20h - h^2)  (h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20 - h) (h) ( h)

\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)

To find maximum/minimum, we have to find first derivative.

(c)

<u>First derivative</u>

\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )

<u>apply chain rule</u>

\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}

<u>Equate the first derivative to zero, that is V'(x) = 0</u>

\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0

\Longrightarrow \sf 40h-3h^2=0

\Longrightarrow \sf h(40-3h)=0

\Longrightarrow \sf h=0, \ 40-3h=0

\Longrightarrow \sf  h=0,\:h=\dfrac{40}{3}<u />

<u>maximum volume:</u>                <u>when h = 40/3</u>

\sf \Longrightarrow max=  \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )

\sf \Longrightarrow maximum= 1241.123 \ cm^3

<u>minimum volume:</u>                 <u>when h = 0</u>

\sf \Longrightarrow min=  \dfrac{1}{3} \pi (0)^2(20-0)

\sf \Longrightarrow minimum=0 \ cm^3

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