word problems: ships' distance, intervals, cost of materials

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bballj228

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Jun 17, 2008
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1: At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 4:00 pm.

2: Find the intervals of increase or decrease
find the local maxiumum and minimum values
find the intervals of concavity and the inflection points.
b(x) = 3x^2/3 - x

3: A rectangular storage container with an open top is to have a volume of 10 m^3. the length of its base is twice the width. Material for the base costs $10 per square meter. Material for the sides cost $6 per square meter. Find the cost of the materials for the cheapest such container.
 
bballj228 said:
2: Find the intervals of increase or decrease
find the local maxiumum and minimum values
find the intervals of concavity and the inflection points.
b(x) = 3x^2/3 - x
Click to expand...
As posted, the function is as follows:

. . . . .\(\displaystyle b(x)\, =\, \frac{3x^2}{3}\, -\, x = x^2\, -\,x\)

...or:

. . . . .\(\displaystyle b(x)\, =\, 3x^{\frac{2}{3}}\, -\, x\)

Did you mean either of these, or something else?

You've taken the first and second derivatives, and gotten... what? You've applied the First and Second Derivative Tests and gotten... what? Where are you stuck?

bballj228 said:
3: A rectangular storage container with an open top is to have a volume of 10 m^3. the length of its base is twice the width. Material for the base costs $10 per square meter. Material for the sides cost $6 per square meter. Find the cost of the materials for the cheapest such container.
Click to expand...
You've drawn the box, picked a variable for the width, written an expression for the length, and picked another variable for the height. You've plugged these and the given volume value into the formula for the volume of a rectangular solid and... then what?

Please be complete. Thank you! :D

Eliz.
 
for number 2. i meant the second one that you wrote. to find the derivative do i multiply 2/3 by 3?
 
For the second one, when taking the derivative (you are correct) you multiply the 3 by (2/3), getting 2. So:

b'(x) = (2/3)3x^[(2/3)-1] - 1

or

b'(x) = 2x^(-1/3) - 1

Now set the derivative equal to zero to determine the intervals on which the function b(x) is increasing or decreasing. For concavity and the point of inflection (POI) you have to take the second derivative:

b''(x) = (-1/3)2x^[(-1/3) - 1] - 0

or

b''(x) = (-2/3)x^(-4/3)

Set that equal to zero and then the intervals where the second derivative, b''(x) is positive, the function, b(x) is concave up (I think) and where the second derivative is negative, the function, b(x) is concave down. I think the POI is where the second derivative changes SIGN and the first derivative equals zero - I'm not really sure about the rules for POI though... Hope that helps and that I didn't make any stupid mistakes!
 
points of inflection occur where the second derivative is equal to 0.
 
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