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uranmaximum [27]

3 years ago

7

Need the answer soon please.

2 answers:

eimsori [14]3 years ago

8 0

I'm not too sure myself (cause Im a learning student myself at the age of 16 and I learn it in a completely different language) I cant really understand c) sorry for my poor english but I think I have gave you enough tips to do c)

kvv77 [185]3 years ago

4 0

Circumference (C) = 2 π r
   122 = 2 π r
61/π = r
Answer to part (b): radius is 61/π ≈ 19.4

Diameter (d) = 2 r
   d = 2 (61/π)
   d = 122/π
Answer to part (a): diameter is 122/π ≈ 38.8

arc length (s) = r · Ф
THETA MUST BE IN RADIANS!
use the unit circle (or use the proportion π/180°) to discover that 30° = π/6

s = (61/π)·(π/6)
   = 61/6
Answer to part (c):arc length is 61/6

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3 years ago

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3 years ago

A cylindrical can, open at the top, is to hold 840 cm3 of liquid. Find the height and radius that minimize the amount of materia

ycow [4]

Answer:

the radius of the base is equal to $x=2\sqrt[3]{\frac{105}{\pi } }$

And the height is equal to:

$y=\frac{840}{\pi(2\sqrt[3]{\frac{105}{\pi}})^{2} }$

Step-by-step explanation:

We write the volume function

$f(x,y)=840=\pi x^{2} y$ where x is the radius of the base and y is the height of the cylinder

$y=\frac{840}{\pi x^{2} }$

The surface of a cylinder is given by

$S(x)=\pi x^{2} +2\pi xy=\pi x^{2} +\frac{1680}{x}$ on the interval from 0 to infinity

We now determine the critical values by differentiating and making the equation equal to 0

$f'(x)=2\pi*x- \frac{1680}{x^{2}} =\frac{2\pi x^{3}-1680}{x^{2} } =0$

This equation have 2 complex solutions and one real solution

$x=2\sqrt[3]{\frac{105}{\pi } }$

For x=0 and infinity the equation goes to infinity also so the radius of the base is equal to $x=2\sqrt[3]{\frac{105}{\pi } }$

And the height is equal to:

$y=\frac{840}{\pi (2\sqrt[3]{\frac{105}{\pi } })^{2} }$

$y=\frac{840}{\pi(2\sqrt[3]{\frac{105}{\pi}})^{2} }$

7 0

3 years ago