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3 years ago

Find the surface area and volume. Answer SA = , V= ___

Mathematics

1 answer:

Tomtit [17]3 years ago

8 0

Answer:

SA = 251.33 yd ^2

V = 301.59 yd ^3

Step-by-step explanation:

Right cylinder 2π = 6.28318530718

Solve for surface area

A =2πrh+2πr2 = 24 x 2π + 2π x 16

A =150.796447372 + 100.530964915

A ≈251.33 yd^2

Right cylinder π = 3.14159265359

Solve for volume

V = πr2h = 16 x 6 = 96π

V = 3.14159265359 x 96

V = 301.592894745

V= 301.59 yd ^3

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3 years ago

A church window consisting of a rectangle topped by a semicircle is to have perimeter p . find the radius of the semicircle if t

Y_Kistochka [10]

Answer:

$r=\dfrac{p}{4+\pi}$

Step-by-step explanation:

Let r be the radius of the semicircle, then 2r is the width of the rectangle. Let y be the length of the rectangle. The perimeter of the window consists of two lengths, one width and length of semicircle, then

$p=2y+2r+\pi r.$

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$y=\dfrac{p-2r-\pi r}{2}.$

The area of the window is

$A=y\cdot 2r+\dfrac{1}{2}\pi r^2.$

Substitute y into the area expression:

$A=\dfrac{p-2r-\pi r}{2}\cdot 2r+\dfrac{1}{2}\pi r^2=pr-2r^2-\pi r^2+\dfrac{1}{2}\pi r^2=pr-2r^2-\dfrac{1}{2}\pi r^2.$

Find the derivative A':

$A'=p-4r-\pi r.$

Equate A' to 0:

$p-4r-\pi r=0\Rightarrow r=\dfrac{p}{4+\pi}.$

When $r$ then $A'>0$ and the function A is increasing, when $r>\dfrac{p}{4+\pi},$ then $A'$ and the function A is decreasing. This means that at point $r=\dfrac{p}{4+\pi}$ the function A takes it maximal value (the area is maximal).