Question

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

Answer 1

`20sqrt2~~28.28" to 2 dec. places"`

Explanation:

`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`

Answer 2

`P_min approx 28.28427`

Explanation:

`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]}