SOLUTION: The profit earned from manufacturing x widgets is .0001x3 - .21x2 + 135x - 2000 dollars. Other contraints dictate that at least 200 widegets but no more than 950 widgets be manufac

Algebra ->  Functions -> SOLUTION: The profit earned from manufacturing x widgets is .0001x3 - .21x2 + 135x - 2000 dollars. Other contraints dictate that at least 200 widegets but no more than 950 widgets be manufac      Log On


   

Question 43154: The profit earned from manufacturing x widgets is .0001x3 - .21x2 + 135x - 2000 dollars. Other contraints dictate that at least 200 widegets but no more than 950 widgets be manufactured.
How many widgets should be produced to earn the maximum profit?
a. 950
b. 17400
c. 500
d. 25500
e. None of the others
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
The profit earned from manufacturing x widgets is
P= .0001x3 - .21x2 + 135x - 2000 ..HOPE YOU KNOW DIFFERENTIATION
DP/DX=0.0001*3X^2-0.21*2X+135=0...FOR P TO HAVE AN EXTREMUM
=0.0003X^2-0.42X+135=0
X=(0.42+ OR - 0.12)/(2*0.0003)
=900 OR 500
D^2P/DX^2=0.0006X-0.42....AT X=900,WE GET...0.0006*900-0.42=0.12....HENCE MINIMUM
AT X=500,WE GET....0.0006*500-0.42=-0.12....HENCE MAXIMUM
SO X=500 IS CORRECT ANSWER...C IS CORRECT
dollars. Other contraints dictate that at least 200 widegets but no more than 950 widgets be manufactured.
How many widgets should be produced to earn the maximum profit?
a. 950
b. 17400
c. 500
d. 25500
e. None of the others