SOLUTION: find the value of x that minimises y=18x^2+(100/x) for positive x. 1. In order to do that first find the derivative dy/dx 2. How many critical numbers does y have, for posit

Algebra ->  Volume -> SOLUTION: find the value of x that minimises y=18x^2+(100/x) for positive x. 1. In order to do that first find the derivative dy/dx 2. How many critical numbers does y have, for posit      Log On


   



Question 1033171: find the value of x that minimises y=18x^2+(100/x) for positive x.

1. In order to do that first find the derivative dy/dx
2. How many critical numbers does y have, for positive x ?
3. What is the nature of the critical number of the previous part?
4. give the exact value of the x that minimises y for positive x?
5. Now give the value approximately as a two place decimal.

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
1. y=18x%5E2%2B100x%5E%28-1%29
dy%2Fdx=36x-100x%5E%28-2%29
dy%2Fdx=36x-100%2Fx%5E%282%29
2.36x-100%2Fx%5E2=0
One
3. Nature?
4. 36x=100%2Fx%5E2
x%5E3=100%2F36
x%5E3=25%2F9
x=%2825%2F9%29%5E%281%2F3%29
5. I leave that to you.