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For a certain company, the cost for producing x items is 40x+300 and the revenue for selling x items is 80x−0.5x2 . 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 ?

User Gourav
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Given that

The cost for producing items is 40x + 300 and the revenue collected is 80x - 0.5x^2

And we have to write an equation for the profit and also we have to find the profit.

Explanation -

The profit is basically the difference between revenue and cost.

So the profit will be


Profit=Revenue-cost

Substituting the values


\begin{gathered} Profit=80x-0.5x^2-(40x+300) \\ Profit=80x-0.5x^2-40x-300 \\ P=40x-0.5x^2-300 \end{gathered}

(a). The profit equation will be, P = 40x - 0.5x^2 - 300

(b). Here the profit given is $300 (P=300) and we have to find the values of x.

Then,


\begin{gathered} 300=40x-0.5x^2-300 \\ 600=40x-0.5x^2 \\ 0.5x^2-40x+600=0 \\ Dividing\text{ by 0.5 we have} \\ x^2-80x+1200=0 \\ x^2-60x-20x+1200=0 \\ x(x-60)-20(x-60)=0 \\ (x-20)(x-60)=0 \\ x=20\text{ and x = 60} \end{gathered}

So the two values of x are 20 and 60.

(c). Now we have to check whether the profit can be $15000 or not.

For this, we have to put P = 15000 in the quadratic equation and if we get a solution for x then it is possible otherwise it is not.

So, substituting P = 15000


\begin{gathered} 15000=40x-0.5x^2-300 \\ 0.5x^2-40x+15300=0 \\ Dividing\text{ by 0.5 in the above equation} \\ x^2-80x+30600=0 \\ \end{gathered}

The solution to this equation is not possible. So Company cannot make a profit of $15000.

Hence the final answers are(a). P = 40x - 0.5x^2 - 300(b). 20 and 60(c). No
User Lisa DeBruine
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