Question

Help please thanks?!

Answer 1

See explanation.

Explanation:

The solid's area is equal to the area of a sphere and a side area of a cylinder, so:

(a)

`A=4pix^2+2pixy` ##

The volume is:

`V=4/3pix^3+pix^2y`

The condition that volume equals `pi/6m^3` allows us to write `y` as the function of `x`:

`4/3pix^3+pix^2y=pi/6`

`4/3x^3+x^2y=1/6`

`8x^3+6x^2y=1`

`y=(1-8x^3)/(6x^2)` (b)

Now we can put the calculated `y` in the area formula:

`A=4pix^2+2pix((1-8x^3)/(6x^2))`

`A=4pix^2+(2pix-16pix^4)/(6x^2)`

`A=4pix^2+pi/(3x)-8/3pix^2`

`A=4/3pix^2+pi/(3x)` (c)

To find the minimum area we have to calculate the derivative `A'`

`A'(x)=8/3pix+pi/3*(-1/x^2)`

`A'(x)=pi/3*(8x-1/x^2)`

`A'(x)=pi/3*((8x^3-1)/x^2)`

`A'(x)=0 iff 8x^3-1=0`

`8x^3=1 => x^3=1/8 => x=1/2`

graph{8x^3-1 [-1, 1, -5, 5]}

From the graph of `A'(x)` we see that the minimum area is reached if `x=1/2`

The minimum area is:

`A(1/2)=4/3pi*(1/2)^2+pi/(3*1/2)`

`A(1/2)=pi/3+2/3pi=pi` (d)

Answer 2

A.
`"Total surface area"="Surface area of cylinder"+"Surface area of two hemispheres"`
`color(white)("Total surface area")="Surface area of cylinder"+"Surface area of a sphere"`
`color(white)("Total surface area")=y(2pix)+4pix^2`
`color(white)("Total surface area")=2pixy+4pix^2`

`A=2pixy+4pix^2`

B.
`pi/6=(4pix^3)/3+ypix^2`

`1/6=(4x^3)/3+yx^2`

`yx^2=1/6-(4x^3)/3`

`y=(1/6-(4x^3)/3)/x^2=1/(6x^2)-(4x)/3`

C.
`A=2pix(1/(6x^2)-(4x)/3)+4pix^2`

`color(white)(A)=(2pix)/(6x^2)-(2pix(4x))/3+4pix^2`

`color(white)(A)=pi/(3x)-(8pix^2)/3+4pix^2`

`color(white)(A)=pi/(3x)+4pix^2-(8pix^2)/3`

`color(white)(A)=pi/(3x)+(12pix^2-8pix^2)/3`

`color(white)(A)=pi/(3x)+(4pix^2)/3`

D.
`A=pi/(3x)+(4pix^2)/3`

`(dA)/(dx)=d/dx[pi/(3x)+(4pix^2)/3]`

`color(white)((dA)/(dx))=d/dx[pi/(3x)]+d/dx[(4pix^2)/3]`

`color(white)((dA)/(dx))=pi/3d/dx[x^(-1)]+(8pix)/3`

`color(white)((dA)/(dx))=pi/3(-x^(-2))+(8pix)/3`

`color(white)((dA)/(dx))=-pi/(3x^2)+(8pix)/3`

`-pi/(3x^2)+(8pix)/3=0`

`(8pix)/3=pi/(3x^2)`

`(8x)/3=1/(3x^2)`

`3x^2(8x)=3`

`8x^3=1`

`x^3=1/8`

`color(white)(x)=1/root(3)(8)`

`color(white)(A)x=1/2`

`A=pi/(3(1/2))+(4pi(1/2)^2)/3`

`color(white)(A)=pi/(3/2)+pi/3`

`color(white)(A)=(2pi)/3+pi/3`

`color(white)(A)=picm^2`