Given that #P=2x+100/x (x>0)#, what is the minimum value of #P#?

#P(x)=2x+100/x=2x+100x^-1#

#rArrP'(x)=2-100x^-2=2-100/(x^2)#

#"we obtain max/min when " P'(x)=0#

#rArr2-100/(x^2)=0larr" multiply through by " x^2#

#rArr2x^2-100=0#

#rArr2(x^2-50)=0#

#rArrx=sqrt50=5sqrt2to x>0#

#"using the "color(blue)"second derivative test"#

#• " if " P''(x)< 0" then maximum"#

#• " if " P''(x)>0" then minimum"#

#P'(x)=2-100x^-2#

#rArrP''(x)=200x^-3=200/(x^3)#

#rArrP''(5sqrt2)=200/(5sqrt2)^3>0#

#rArrP(x)" is a minimum when " x=5sqrt2#

#P(5sqrt2)=(2xx5sqrt2)+100/(5sqrt2)#

#color(white)(P(5sqrt2))=10sqrt2+(100sqrt2)/10#

#color(white)(P(5sqrt2))=10sqrt2+10sqrt2=20sqrt2~~28.28#

#P= 2x+100/x#

#(dP)/dx= 2-100/x^2# (Power rule)

For a maximum or minimum value of #P#, #(dP)/dx =0#

#2-100/x^2 = 0#

#100/x^2 =2#

#x^2 = 100/2 =50#

#x=+-sqrt50#

We are told that #x>0 -> x=sqrt50#

To test for a local maximum or minimum P:

#(d^2p)/dx^2 = 0+200/x^3# which is #>0# for #x>0#

Hence, #P_min = P(sqrt50) approx P(7.07107) approx 28.28427 #

The local minimum can be seen on the graph of #P# below.

graph{2x+100/x [-150.1, 150.1, -75, 75.1]}