Answer:
Profit function: P(x) = -0.5x^2 + 40x - 300
profit of $50: x = 10 and x = 70
NOT possible to make a profit of $2,500, because maximum profit is $500
Step-by-step explanation:
(Assuming the correct revenue function is 90x−0.5x^2)
The cost function is given by:

And the revenue function is given by:

The profit function is given by the revenue minus the cost, so we have:



To find the points where the profit is $50, we use P(x) = 50 and then find the values of x:



Using Bhaskara's formula, we have:



So the values of x that give a profit of $50 are x = 10 and x = 70
To find if it's possible to make a profit of $2,500, we need to find the maximum profit, that is, the maximum of the function P(x).
The maximum value of P(x) is in the vertex. The x-coordinate of the vertex is given by:

Using this value of x, we can find the maximum profit:

The maximum profit is $500, so it is NOT possible to make a profit of $2,500.