Stationary Points

  • Thread starter jcrane
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A local inventor has developed a new camera technology that uses special tubes to
improve the quality of the pictures it takes. To produce, each camera will cost
$1,000 of fixed costs, $25 for every tube used, and $400 divided by the number of
tubes used in each camera. The camera sells for $2,025.


a) Write the function P(x) for the profit on a camera, where x represents the number
of tubes used in each camera.
b) What obvious restriction must be placed on x?
c) Use the function in a) to determine the number of tubes that will maximize profit.
d) Show that P(x) is maximized at the stationery point.



Attempted Solution

a) P(x) = 1025 - 25x - 400/x

b) x > 0

c) i managed to get x = +/- 4 . Therefore 10 tubes would maximize profit .

d) *confused on this one
 

Answers and Replies

  • #2
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A local inventor has developed a new camera technology that uses special tubes to
improve the quality of the pictures it takes. To produce, each camera will cost
$1,000 of fixed costs, $25 for every tube used, and $400 divided by the number of
tubes used in each camera. The camera sells for $2,025.


a) Write the function P(x) for the profit on a camera, where x represents the number
of tubes used in each camera.
b) What obvious restriction must be placed on x?
c) Use the function in a) to determine the number of tubes that will maximize profit.
d) Show that P(x) is maximized at the stationery point.



Attempted Solution

a) P(x) = 1025 - 25x - 400/x

b) x > 0

c) i managed to get x = +/- 4 . Therefore 10 tubes would maximize profit .
Isn't x the number of tubes used to make a camera? You just found that x = 4 (x = -4 is not in the domain.).
d) *confused on this one
You have a maximum if P'(x) > 0 for x to the left of the stationary point, and P'(x) < 0 for x to the right of the stationary point. (Stationery is paper and envelopes and such.)
 
  • #3
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Sorry, i just copied the question from the assignment website ..
 

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