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hram777 [196]
2 years ago
9

Work out the surface area of a cylinder when the height = 18cm and the volume = 1715cm cubed

Mathematics
2 answers:
lara [203]2 years ago
8 0

Answer:

813.4 cm² (nearest tenth)

Step-by-step explanation:

<u>Volume of a cylinder</u>

\sf V=\pi r^2 h \quad\textsf{(where r is the radius and h is the height)}

Given:

  • h = 18cm
  • V = 1715 cm³

Use the Volume of a Cylinder formula and the given values to find the <u>radius of the cylinder</u>:

\implies \sf 1715=\pi r^2 (18)

\implies \sf r^2=\dfrac{1715}{18 \pi}

\implies \sf r=\sqrt{\dfrac{1715}{18 \pi}

<u>Surface Area of a Cylinder</u>

\sf SA=2 \pi r^2 + 2 \pi r h \quad\textsf{(where r is the radius and h is the height)}

Substitute the given value of h and the found value of r into the formula and solve for SA:

\implies \sf SA=2 \pi \left(\sqrt{\dfrac{1715}{18 \pi}\right)^2 + 2 \pi \left(\sqrt{\dfrac{1715}{18 \pi}\right)(18)

\implies \sf SA=2 \pi \left(\dfrac{1715}{18 \pi} \right) + 36 \pi \left(\sqrt{\dfrac{1715}{18 \pi}\right)

\implies \sf SA=\dfrac{1715}{9} + 36 \pi \left(\sqrt{\dfrac{1715}{18 \pi}\right)

\implies \sf SA=813.3908956...

Therefore, the surface area of the cylinder is 813.4 cm² (nearest tenth)

zubka84 [21]2 years ago
7 0

We need radius

  • πr²h=1715
  • πr²(18)=1715
  • r²=30.3
  • r=5.5cm

Now

LSA

  • 2πrh
  • 2π(5.5)(18)
  • 11(18)(3.14)
  • 198(3.14)
  • 622.72cm²
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Step-by-step explanation:

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4 0
2 years ago
Please see attachment
Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   x = \frac{1}{\sqrt{2} }  

Step-by-step explanation:

<u>Step(i)</u>:-

Given function

                       f(x) = \frac{-x}{2x^{2} +1}     ...(i)

Differentiating equation (i) with respective to 'x'

                     f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2}  }   ...(ii)

                    f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2}  }

Equating Zero

                   f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2}  } = 0

                 \frac{2x^{2}-1}{(2x^{2}+1)^{2}  } = 0

                2 x^{2}-1 = 0

               2 x^{2} = 1

             x^{2}  = \frac{1}{2}

             x = \frac{-1}{\sqrt{2} }  , x = \frac{1}{\sqrt{2} }

<u><em>Step(ii):</em></u>-

Again Differentiating equation (ii) with respective to 'x'

f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4}  }

put

      x = \frac{1}{\sqrt{2} }

f^{ll} (x) > 0

The absolute minimum value at   x = \frac{1}{\sqrt{2} }

<u><em>Step(iii):</em></u>-

The value of absolute minimum value

                         f(x) = \frac{-x}{2x^{2} +1}

                       f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}

         on calculation we get

The value of absolute minimum value = - 0.3536      

<u><em>Final answer</em></u>:-

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   x = \frac{1}{\sqrt{2} }    

3 0
2 years ago
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