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Remember that
The Profit function P() is the difference between the revenue function R(x) and the total cost function C()
In this problem
the revenue function is equal to
R(p)=x*p
where
x=6,000-30p
R(p)=(6,000-30p)p
R(p)=6,000p-30p^2
C(x)=72,000+60x
C(p)=72,000+60(6,000-30p)
C(p)=72,000+360,000-1,800p
C(p)=432,000-1,800p
Part A
Break even
R(p)=C(p)
substitute
6,000p-30p^2=432,000-1,800p
solve the system by graphing
using a graphing tool
the values of p are
p=$80 and p=$180
Part B
Maximum revenue
R(p)=6,000p-30p^2
this is a vertical parabola open a downward
the vertex is a maximum
the y-coordinate of the vertex is the maximum revenue
using a graphing tool
the vertex is the point (100,300,000)
therefore
the maximum revenue is $300,000 For a p=$100
Part C
Find the price that produces the maximum profit.
P(p)=R(p)-C(p)
P(p)=(6,000p-30p^2)-(432,000-1,800p)
P(p)=-30p^2+7,800p-432,000
The maximum profit is the y-coordinate of the vertex
using a graphing tool
the vertex is (130,75,000)
therefore
The maximum profit is $75,000
