Answer:
A)
[tex]\begin{equation*} C^{\prime}(q)=110 \end{equation*}[/tex]B)
[tex]\begin{equation*} R(q)=180q-\frac{q^2}{50} \end{equation*}[/tex]C)
[tex]\begin{equation*} R^{\prime}(q)=180-\frac{q}{25} \end{equation*}[/tex]D)
[tex]\begin{equation*} R^{\prime}(1100)=136 \end{equation*}[/tex]E)
[tex]\begin{equation*} P(q)=-\frac{q^2}{50}+70q-72000 \end{equation*}[/tex]F)
[tex]\begin{equation*} P^{\prime}(1100)=26 \end{equation*}[/tex]Explanation:
Given:
[tex]\begin{gathered} p=180-\frac{q}{50} \\ C(q)=72000+110q \end{gathered}[/tex]where q = the number of microwaves that can be sold at a price of p dollars per unit
C(q) = the total cost (in dollars) of producing q units.
A) To find the marginal cost, C'(q), we'll go ahead and take the derivative of the total cost as seen below;
[tex]\begin{gathered} C(q)=72000+110q \\ C^{\prime}(q)=0+110 \\ \therefore C^{\prime}(q)=110 \end{gathered}[/tex]So the marginal cost, C'(q) = 110
B) We'll go ahead and determine the revenue function, R(q), by multiplying the price, p, by the quantity, q, as seen below;
[tex]\begin{gathered} R(q)=p*q=(180-\frac{q}{50})q=180q-\frac{q^2}{50} \\ \therefore R(q)=180q-\frac{q^2}{50} \end{gathered}[/tex]C) We'll go ahead and determine the marginal revenue function, R'(q), by taking the derivative of the revenue function, R(q);
[tex]\begin{gathered} \begin{equation*} R(q)=180q-\frac{q^2}{50} \end{equation*} \\ R^{\prime}(q)=180-\frac{2q^{2-1}}{50}=180-\frac{q}{25} \\ \therefore R^{\prime}(q)=180-\frac{q}{25} \end{gathered}[/tex]D) To evaluate the marginal revenue function at q = 1100, all we need to do is substitute the q with 1100 in R'(q) and simplify;
[tex]\begin{gathered} \begin{equation*} R^{\prime}(q)=180-\frac{q}{25} \end{equation*} \\ R^{\prime}(1100)=180-\frac{1100}{25}=180-44=136 \\ \therefore R^{\prime}(1100)=136 \end{gathered}[/tex]Therefore, R'(1100) is 136
E) To find the profit function, P(q), we have to subtract the total cost, C(q), from the revenue cost, R(q);
[tex]\begin{gathered} P(q)=R(q)-C(q) \\ =(180q-\frac{q^2}{50})-(72,000+110q) \\ =180q-\frac{q^2}{50}-72000-110q \\ =-\frac{q^2}{50}+180q-110q-72000 \\ =-\frac{q^2}{50}+70q-72000 \\ \therefore P(q)=-\frac{q^2}{50}+70q-72000 \end{gathered}[/tex]F) To Evaluate the marginal profit function at q = 1100, we have to first determine the marginal profit, P'(q), by taking the derivative of the profit function, P(x);
[tex]\begin{gathered} \begin{equation*} P(q)=-\frac{q^2}{50}+70q-72000 \end{equation*} \\ P^{\prime}(q)=-\frac{2q^{2-1}}{50}+70(1*q^{1-1})-0=-2q^{50}+70q^0=-\frac{q}{25}+70 \\ \therefore P^{\prime}(q)=-\frac{q}{25}+70 \end{gathered}[/tex]We can now go ahead and find P'(1100) as seen below;
[tex]\begin{gathered} P^{\prime}(1100)=-\frac{1100}{25}+70=-44+70=26 \\ \therefore P^{\prime}(1100)=26 \end{gathered}[/tex]So P'(1100) = 26
