quadratic polynomial question

Question

For a certain company, the cost for producing x items is 60x+300 and the revenue for selling x items is 100x−0.5x. The profit that the company makes is how much it takes in (revenue) minus how much it spends (cost). In economic models, one typically assumes that a company wants to maximize its profit, or at least wants to make a profit!

Part a: Set up an expression for the profit from producing and selling x items. We assume that the company sells all of the items that it produces. (Hint: it is a quadratic polynomial.)

Part b: Find two values of x that will create a profit of $300.

Part c: Is it possible for the company to make a profit of $15,000?

Answer 1

I think I've seen the question before. In this situation, since the question mentions that the combined expression for profit is quadratic, one of the "x"s in the equations given should be squared, and I would assume it is the x in the selling equation, since the "x" arent combined and I would think that a coefficient of 0.5 is most probable to be in front of the x^2. Do correct me if the question has the x^2 placed somewhere else.

Now Profit = Revenue - Cost

a) Therefore Profit = (100x - 0.5x^2) - (60x + 300) = -.5x^2 + 100x - 60x - 300 = -0.5x^2 + 40x - 300

b) Now to find the two values that produce a profit of $300, we will input 300 into the profit location and solve for x from our new equation; 300 = -0.5x^2 + 40x - 300. You can use a calculator or use the quadratic formula to find the two x values. Try it out, if you can't get it reply to my comment and I will help further.

c) This is basically similar. Plug in 15,000 for Profit in our equation from a) and try to see if you can find any values for x. If you cannot obtain any real values for x, or obtain negative values for x(which means you need NEGATIVE number of units to be produced, which is not possible), then it means attaining that profit is not possible. If you do find a real number, positive value for x, then it is possible. Reply to my comment if any further help is still required.