Respuesta :
Answer:
At Maximum point;
x(max) = 50
p(max) = $100
Maximum revenue = $5,000
Step-by-step explanation:
The price function is;
p(x) = 200 − 2x
where
p is the price (in dollars) at which exactly x trucks will be rented per day.
The revenue function R(x) can be written as;
R(x) = p(x) × x
Substituting p(x) equation;
R(x) = (200-2x)x
R(x) = 200x-2x^2 ........1
To maximize R(x), at maximum point dR/dx = 0
differentiating equation 1;
dR/dx = 200 - 4x = 0
4x = 200
x = 200/4
x = 50
Substituting x = 50 into p(x)
p(50) = 200 - 2(50) = $100
p = $100
Maximum revenue is;
R = p × x = $100×50
R = $5,000
At the level of the maximum point, the value of x(max) should be 50 and the value of p(max) should be $100, Also, Maximum revenue = $5,000.
Calculation of the maximum value:
Since
The price function is;
p(x) = 200 − 2x
Here,
p refer to the price (in dollars) at which exactly x trucks should be rented per day.
Now
The revenue function R(x) should be
R(x) = p(x) × x
Now
R(x) = (200-2x)x
R(x) = 200x-2x^2 ........1
Now
To maximize R(x), at maximum point dR/dx = 0
So,
dR/dx = 200 - 4x = 0
4x = 200
x = 200/4
x = 50
Now
Substituting x = 50 into p(x)p(50) = 200 - 2(50) = $100
p = $100
Now
Maximum revenue is;
R = p × x = $100×50
R = $5,000
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