In order to find the expression for the profit, let's subtract the revenue and the cost:
[tex]\begin{gathered} Profit=Revenue-Cost\\ \\ Profit=85x-0.5x^2-(45x+300)\\ \\ Profit=-0.5x^2+40x-300 \end{gathered}[/tex]Now, to find two values of x that create a profit of 300, let's use Profit = 300 and solve for x:
[tex]\begin{gathered} 300=-0.5x^2+40x-300\\ \\ -0.5x^2+40x-600=0\\ \\ x^2-80x+1200=0\\ \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \\ x=\frac{80\pm\sqrt{6400-4800}}{2}\\ \\ x_1=\frac{80+40}{2}=\frac{120}{2}=60\\ \\ x_2=\frac{80-40}{2}=\frac{40}{2}=20 \end{gathered}[/tex]Therefore the two values are x = 20 and x = 60.
Now, let's find if the profit can be $15,000:
[tex]\begin{gathered} 15000=-0.5x^2+40x-300\\ \\ -0.5x^2+40x-15300=0\\ \\ x^2-80x+30600=0\\ \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \\ x=\frac{80\pm\sqrt{6400-122400}}{2}\\ \\ x=40\pm\frac{\sqrt{-116000}}{2}\\ \\ x=40\pm170.29i \end{gathered}[/tex]Since the solutions are complex numbers, therefore the answer is NO.
