SOLUTION: Demand Equation The price (in dollars) and the quantity x sold of a certain product obey the demand equation p= -1/10x+150, 0≤x≤1500 Revenue is x*p. (a) Express the

Algebra ->  Rational-functions -> SOLUTION: Demand Equation The price (in dollars) and the quantity x sold of a certain product obey the demand equation p= -1/10x+150, 0≤x≤1500 Revenue is x*p. (a) Express the      Log On


   

Question 767975: Demand Equation The price (in dollars) and the quantity x sold of a certain product obey the demand equation p= -1/10x+150, 0≤x≤1500
Revenue is x*p.
(a) Express the revenue R as a function of x.
(b) What quantity x maximizes revenue? What is the maximum revenue?
***PLEASE SHOW STEPS!!
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
++p=+%28-1%2F10%29%2Ax%2B150+, 0≤x≤1500
+R+=+p%2Ax+
+R+=+%28+%28-1%2F10%29%2Ax+%2B+150+%29%2Ax+
+R+=+%28+-1%2F10+%29%2Ax%5E2+%2B+150x+
This is a parabola with a maximum since
the coefficient of the +x%5E2+ term is minus.
Compare the equation with the general form:
+y+=+ax%5E2+%2B+b%2Ax+%2B+c+
+a+=+-1%2F10+
+b+=+150+
+c+=+0+
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The x-coordinate of the maximum is at
+-b%2F%282a%29+=+-150+%2F+%28+2%2A%28-1%2F10%29%29+
+-b%2F%282a%29+=+150+%2F+%281%2F5%29+
+-b%2F%282a%29+=+750+
This is the value of x that maximizes revenue
Plug this back into the equation
+R%5Bmax%5D+=+%28+-1%2F10+%29%2Ax%5E2+%2B+150x+
+R%5Bmax%5D+=+%28+-1%2F10+%29%2A750%5E2+%2B+150%2A750+
+R%5Bmax%5D+=+-562500%2F10+%2B+112500+
+R%5Bmax%5D+=+-56250+%2B+112500+
+R%5Bmax%5D+=+56250+
The maximum revenue is $56,250