Answer:
$100 per phone, and the maximum revenue is $5,200.
To find the selling price that will give the maximum revenue, you need to determine the price (xx) that maximizes the revenue function R(x)=−0.52x2+104xR(x)=−0.52x2+104x.
The maximum revenue occurs at the vertex of the quadratic function, and the x-coordinate of the vertex can be found using the formula x=−b2ax=−2ab, where aa is the coefficient of the quadratic term (-0.52) and bb is the coefficient of the linear term (104).
In this case:
a=−0.52a=−0.52
b=104b=104
Now, calculate xx:
x=−1042(−0.52)x=−2(−0.52)104
x=−104−1.04x=−−1.04104
x=100x=100
So, the selling price that will give the maximum revenue is $100 per phone.
To find the maximum revenue, substitute this selling price back into the revenue function R(x)R(x):
R(100)=−0.52(100)2+104(100)R(100)=−0.52(100)2+104(100)
R(100)=−0.52(10,000)+10,400R(100)=−0.52(10,000)+10,400
R(100)=−5,200+10,400R(100)=−5,200+10,400
R(100)=5,200R(100)=5,200
The maximum revenue is $5,200.
So, the selling price that will give the maximum revenue is $100 per phone, and the maximum revenue is $5,200.