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We have a product that has a revenue function R(p) and an expense function E(p):
[tex]\begin{gathered} R(p)=-150p^2+3500p \\ E(p)=-700p+27000 \end{gathered}[/tex]a) The profit function will be the difference between the revenue and the expense, for any value of p, so it will be a also a function. It can be calculated as:
[tex]\begin{gathered} P(p)=R(p)-E(p) \\ P(p)=(-150p^2+3500p)-(-700p+27000) \\ P(p)=-150p^2+3500p+700p-27000 \\ P(p)=-150p^2+4200p-27000 \end{gathered}[/tex]b) The breakeven point happens when P(p) = 0, meaning that, at that price, the profit is 0 or the revenue is equal to the expense.
We can find this breakeven points as the roots of the profit function. We will start by simplyfing the coefficients as:
[tex]\begin{gathered} P(p)=0 \\ -150p^2+4200p-27000=0 \\ \frac{-150p^2+4200p-27000}{-300}=\frac{0}{-300} \\ \frac{1}{2}p^2-14p+90=0 \end{gathered}[/tex]Then, we calculate the roots as:
[tex]\begin{gathered} p=\frac{-(-14)\pm\sqrt[]{(-14)^2-4\cdot\frac{1}{2}\cdot90}}{2\cdot\frac{1}{2}} \\ p=14\pm\sqrt[]{196-180} \\ p=14\pm\sqrt[]{16} \\ p=14\pm4 \\ p_1=14-4=10 \\ p_2=14+4=18 \end{gathered}[/tex]Then, the two breakeven points are at p=10 and p=18. We can see them in a graph like:
c) We have to find the maximum profit. To do this we first have to find at which price this maximum profit happens.
We have two ways:
1) As we know that P(p) is a quadratic function and we know its roots, we know that the maximum profit will be represented by the vertex.
2) We can derive the profit function and equal to 0. This allows us to find a extreme in the function, where the slope of the tangent line is 0.
We will use the first method.
The x-coordinate of the vertex of P(p) can be calculated from the coefficients as:
