Help please thanks?!

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2 Answers
Feb 2, 2018

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)

Feb 2, 2018

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#