SOLUTION: Find the value of x that minimises y = 16x^2 + 1800/x for positive x. So first I need to differentiate, which would give y' = 32x

Click here to see ALL problems on Finance

Question 990242: Find the value of x that minimises
y = 16x^2 + 1800/x
for positive x.
So first I need to differentiate, which would give y' = 32x - 1800/x^2
I am stuck with the next 4 parts..
How many critical numbers does y have, for positive x?
Recall that a critical number of a function is a value of x for which the derivative of the function is zero or doesn't exist.
What is the nature of the critical number of the previous part?
Enter m for minimum, M for maximum or i for a horizontal point of inflection.
Give the exact value of the x that minimises y for positive x.
x (at minimum) =
Now give the value approximately as a 2 decimal place decimal. Now, give the approximate value (as a 2 decimal place decimal) of the x that minimises y for positive x.
x (at minimum) ≈

THANK YOU
Answer by josgarithmetic(39631) (Show Source):
You can put this solution on YOUR website!
Where is the derivative equal to zero?
Simplify that derivative and then solve this.

The factor of 8 can probably not be needed because you are looking for the NUMERATOR to be zero; and for the same reason, the x in the numerator is not important for this.
You now just need to solve for .

SOLUTION: Find the value of x that minimises y = 16x^2 + 1800/x for positive x. So first I need to differentiate, which would give y' = 32x - 1800/x^2 I am stuck with the next