Question 1118294
Look for minimum AREA for the minimum cost.


h, height
r, radius
A, area


{{{system(h*pi*r^2=16pi,A=2*pi*r^2+h*2pi*r)}}}

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{{{hr^2=16}}}

{{{h=16/r^2}}}

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{{{A=2pi*r^2+(16/r^2)(2pi*r)}}}
{{{A=2pi*r^2+32pi/r}}}------------------(technology is showing minimum at about r=2)

{{{dA/dr=4pi*r+(32pi)(-1)(r^(-2))}}}


{{{dA/dr=4pi*r-32pi/r^2}}}
{{{common_denominator_r^2}}}

{{{dA/dr=(4pir^3-32pi)/r^2}}}

Need the NUMERATOR to be 0.

{{{4pi*r^3-32pi=0}}}

{{{4pi*r^3=32pi}}}

{{{r^3=8}}}

{{{highlight(r=2)}}}
.

With this r value found for minimum A, the value for h, minimum A, can be evaluated, and also the corresponding cost.