The profit equation is:
p(x) = -0.5*x^2 + 70x - 200
The domain is:
x ∈ [0, 140], such that x ∈ Z
and the profit is larger for 70 items.
We know that:
Cost equation:
c(x) = 50*x + 200
revenue equation:
r(x) = -0.5*(x - 120)^2 + 7,200
The maximum capacity is 140
Then x can be any value in the range [0, 140]
We want to find the profit equation, remember that:
profit = revenue - cost
Then the profit equation is:
p(x) = r(x) - c(x)
p(x) = ( -0.5*(x - 120)^2 + 7,200) - ( 50*x + 200)
Now we can simplify this:
p(x) = -0.5*(x - 120)^2 + 7,200 - 50x - 200
p(x) = -0.5*(x - 120)^2 + 7,000 - 50x
p(x) = -0.5*(x^2 - 2*120*x + 120^2) + 7,000 - 50x
p(x) = -0.5*x^2 + 120x - 7,200 + 7,000 - 50x
p(x) = -0.5*x^2 + 70x - 200
Now we want to answer, what is the domain of p(x)?
The domain is the set of the possible inputs of the function.
Remember that x is in the range [0, 140], such that x should be a whole number, so we also need to add x ∈ Z
then:
x ∈ Z ∧ x ∈ [0, 140]
Then that is the domain of the profit function.
Now we want to see the profit for 70 and 80 items, to do it, just evaluate p(x) in these values:
70 items:
p(70) = -0.5*70^2 + 70*70 - 200 = 2,250
p(80) = -0.5*80^2 + 70*80 - 200 = 2,200
So the profit is smaller for the larger number of items, why does this happen?
You can see that the profit equation is a quadratic equation with a negative leading coefficient, so, as the value of x increases after a given point (the vertex of the quadratic) the profit will start to decrease.
If you want to learn more about quadratic functions, you can read:
https://brainly.com/question/22889366
