SOLUTION: Solve the following problem. Ace manufacturing has determined that the cost of labor for producing transmissions is 0.3x^2+400x+550 dollars, while the cost of materials is 0.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Solve the following problem. Ace manufacturing has determined that the cost of labor for producing transmissions is 0.3x^2+400x+550 dollars, while the cost of materials is 0.      Log On


   

Question 20107: Solve the following problem.
Ace manufacturing has determined that the cost of labor for producing transmissions is 0.3x^2+400x+550 dollars, while the cost of materials is 0.1x^2+50x+800 dollars.
a. Write a polynomial that represents the total cost of materials and labor for producing x transmissions.
b. Evaluate the total cost polynomial for x=500.
c. Find the cost of labor for 500 transmissions and the cost of materials for 500 transmissions.
Thanks
Answer by pwac(253) About Me  (Show Source):
You can put this solution on YOUR website!
a)add the two expressions0.3x%5E2%2B400x%2B550+0.1x%5E2%2B50x%2B800
= 0.4x%5E2%2B450x%2B1350
b)substitute x=500 into above expression
%280.4%28500%29%5E2%29%2B450%28500%29%2B1350
=100 000+225 000+1350
=$326350
c)Labour=%280.3%28500%29%5E2%29%2B400%28500%29%2B550
=$275550
Materials=%280.1%28500%29%5E2%29%2B50%28500%29%2B800
=$50800.
Or simply Total cost-Labour=Materials
Pete