SOLUTION: Find the value of x that minimises y = 16x^2 + 1800/x for positive x So I know I need to differentiate, but I am not sure where to go from there? 32x

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Question 993477: Find the value of x that minimises
y = 16x^2 + 1800/x
for positive x
So I know I need to differentiate, but I am not sure where to go from there?
32x - (1800/x^2)
I need:
How many critical numbers does y have, for positive x?
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 rothauserc(4718) (Show Source):
You can put this solution on YOUR website!
y = 16x^2 + 1800/x
dy/dx = 32x -(1800/x^2)
to find min x for positive x set dy/dx = 0
32x -(1800/x^2) = 0
multiply both sides of = by x^2
32x^3 - 1800 = 0
32x^3 = 1800
x = (15/2)^(2/3)
for positive x, x is a local minimum
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x(at minimum) = (15/2)^(2/3)
x(at minimum) = 3.831547162 approx 3.83
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here is the graph of y = 16x^2 + 1800/x to help visualize this problem

SOLUTION: Find the value of x that minimises y = 16x^2 + 1800/x for positive x So I know I need to differentiate, but I am not sure where to go from there? 32x - (1800/x^2) I