SOLUTION: Total profit=total revenue-total cost, P(x)=r(x)-C(x) Where x is the nyumber of units sold, find the maximum profit and the number of units that must be sold in order to yield th

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Total profit=total revenue-total cost, P(x)=r(x)-C(x) Where x is the nyumber of units sold, find the maximum profit and the number of units that must be sold in order to yield th      Log On


   



Question 451173: Total profit=total revenue-total cost,
P(x)=r(x)-C(x)
Where x is the nyumber of units sold, find the maximum profit and the number of units that must be sold in order to yield the maximum profit for
R(x)=5x, C(x)=0.001x^2+1.2x+60

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+p%28x%29++=+r%28x%29+-+c%28x%29+
+p%28x%29+=+5x+-+%28.001x%5E2+%2B+1.2x+%2B+60+%29+
+p%28x%29+=+5x+-+.001x%5E2+-+1.2x+-+60+
+p%28x%29+=+-.001x%5E2+%2B+3.8x+-+60+
I need to find the max x of the function +p%28x%29+
This occurs at +x+=+-b%2F%282a%29+ when the equation
has the form +f%28x%29+=+ax%5E2+%2B+bx+%2B+c+
+-b%2F%282a%29+=+-3.8%2F%28-.002%29+
+-b%2F%282a%29+=+1900+
When 1900 units are sold, the profit is a maximum
here's the graph:
+graph%28+600%2C+600%2C+-300%2C+4000%2C+-300%2C+4000%2C+-.001x%5E2+%2B+3.8x+-+60%29+