Maths question integration possibly?

Question

can you please help me with this question, and show as much working as you can be bothered to type?

P= 2*pi [r^2 + (128/r)]

What is the value of R, so that P has its minimum value?

thanks for your help in advance! :)

Answer 1

If you can graph this function you will see that for r < 0 the function rises to +infinity as r--> -infinity and drops to -infinity as r--> 0 from the left.  For r > 0 the function experiences a local minimum and I'm guessing that's the minimum you'd like to find via calculus.  When

r = 0 the function blows up since you're dividing by 0.    

The derivative of a function gives the slope of the function at a selected point.  You may recall that when the derivative of a function is 0 (means the slope of the function is 0), the function may have a local minimum or maximum.  

Step one then, is to get the derivative of P (dP/dr)

Step two:  dP/dr = 0 to find r such that the function's slope is 0.

Step three:  Solve for r (I got r = 4, a triple repeated root)

Step four:  find the value of the minimum (if you want) by plugging in r = 4 into the --original-- function

Answer 2

You meant to type a small 'r' right?

Find the first derivative.

P=2π(r²+(128/r))

P=2πr²+(256π/r)

P'=4πr-(256π/r²)

P''=4π+(512πr)/r^4

4πr-(256π/r²)=0

(4πr³-256π)/r²=0

4πr³-256π=0

4πr³=256π

r³=64

r=4

P''(4)=4π+((512(4))/4^4)

This is positive so a minimum occurs at r=4

Value is r=4 and the minimum is P=96π

Answer: r=4