Respuesta :
Answer:
Profit function: [tex]P(x) = -0.5x^2 + 40x - 300[/tex]
Values where the profit is $50: [tex]x_1 = 70[/tex], [tex]x_2 = 10[/tex]
It is NOT possible to make a profit of $2,500, the maximum profit is $500.
Step-by-step explanation:
To find the profit function P(x), we just need to calculate the revenue function R(x) minus the cost function C(x).
In our case, the cost function is:
[tex]C(x) = 50x+300[/tex]
And the revenue function is:
[tex]R(x) = 90x - 0.5x^2[/tex]
(I'm assuming this is the correct revenue and the question is missing an 'x' after 0.5 and before '^2')
So the profit function is:
[tex]P(x) = R(x) - C(x)[/tex]
[tex]P(x) = 90x - 0.5x^2 - 50x - 300[/tex]
[tex]P(x) = -0.5x^2 + 40x - 300[/tex]
To find the values of x that give a profit of $50, we use P(x) = 50 and then find the values of x:
[tex]50 = -0.5x^2 + 40x - 300[/tex]
[tex]-0.5x^2 + 40x - 350 = 0[/tex]
[tex]x^2 - 80x + 700 = 0[/tex]
Using Bhaskara's formula, we have:
[tex]\Delta = b^2 - 4ac = (-80)^2 - 4*1*700 = 3600[/tex]
[tex]x_1 = (-b + \sqrt{\Delta})/2a = (80+60)/2 = 70[/tex]
[tex]x_2 = (-b - \sqrt{\Delta})/2a = (80-60)/2 = 10[/tex]
So the values of x area 10 and 70 items.
To find if it's possible to make a profit of $2,500, let's find the maximum value of our profit function.
We can find the value of x that gives the maximum profit finding the x of the vertex using this formula:
[tex]x_v = -b/2a = 80/2 = 40[/tex]
Now, using this value of x in the profit equation, we have the maximum profit:
[tex]P(40) = -0.5*(40)^2 + 40*40 - 300 = 500[/tex]
The maximum profit of the company is $500, so it is not possible to make a profit of $2,500.