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For a certain company, the cost function for producing x items is C(x)=50x+200 and the revenue function for selling x items is R(x)=−0.5(x−120)2+7,200. The maximum capacity of the company is 140 items.

Assuming that the company sells all that it produces, what is the profit function?
P(x)=
Preview Change entry mode .

Hint: Profit = Revenue - Cost as we examined in Discussion 3.

What is the domain of P(x)?
Hint: Does calculating P(x) make sense when x=−10 or x=1,000?

The company can choose to produce either 70 or 80 items. What is their profit for each case, and which level of production should they choose?
Profit when producing 70 items =
Number


Profit when producing 80 items =
Number


Can you explain, from our model, why the company makes less profit when producing 10 more units?

Sagot :

facundo3141592

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

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