SOLUTION: http://i.imgur.com/OQ38hve.png Here is picture if anything came out wrong in copy/paste Find the value of x that minimises y = 10 x2 + 1900 x for positive x. 2.1

Question 970316: http://i.imgur.com/OQ38hve.png
Here is picture if anything came out wrong in copy/paste

Find the value of x that minimises
y = 10 x2 + 1900
x
for positive x.

2.1 (1 mark)

In order to do that first find the derivative dy/dx.
dy/dx =
Your last answer was:
20 x−1900 x−2
Your answer is correct.
The mark for your last attempt was 1.00
You previously made one incorrect attempt, and were penalised 0.10, giving an overall mark of 0.90.
2.2 (1 mark)

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.
Answer:
Your last answer was:
4.56
Wrong type
Your answer should have type
integer
but it does not.
2.3 (1 mark)

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.
Answer:
You have not attempted this yet
2.4 (1 mark)

Give the exact value of the x that minimises y for positive x.
x (at minimum) =
You have not attempted this yet
2.5 (1 mark)

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) ≈
You have not attempted this yet
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
2.1) dy/dx = -1900x^(-2) + 20x
2.2) -1900x^(-2) + 20x = 0
(20(x^3 - 95)) / x^2 = 0
for x > 0
x^3 - 95 = 0
x = (95)^(1/3)
answer is 1 solution
2.3) m (minimum)
2.4) x = (95)^(1/3) = 4.562902635
2.5) x(at minimum) = 4.56

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