Question

The price-demand and cost functions for the production of microwaves are given asP= 180 - q/50and C(q) = 72000 + 110g,where q is the number of microwaves that can be sold at a price of p dollars per unit and C(q) is the total cost (in dollars) of producing q units.(C) Find the marginal revenue function in terms of q.R'(q) =(D) Evaluate the marginal revenue function at q=1100.R'(1100) =(E) Find the profit function in terms of q.P(q)(F) Evaluate the marginal proft function at q = 1100.P'(1100)

Answer 1

A)

`\begin{equation*} C^(\prime)(q)=110 \end{equation*}`

B)

`\begin{equation*} R(q)=180q-(q^2)/(50) \end{equation*}`

C)

`\begin{equation*} R^(\prime)(q)=180-(q)/(25) \end{equation*}`

D)

`\begin{equation*} R^(\prime)(1100)=136 \end{equation*}`

E)

`\begin{equation*} P(q)=-(q^2)/(50)+70q-72000 \end{equation*}`

F)

`\begin{equation*} P^(\prime)(1100)=26 \end{equation*}`

Step-by-step explanation:

Given:

`\begin{gathered} p=180-(q)/(50) \\ C(q)=72000+110q \end{gathered}`

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;

`\begin{gathered} C(q)=72000+110q \\ C^(\prime)(q)=0+110 \\ \therefore C^(\prime)(q)=110 \end{gathered}`

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;

`\begin{gathered} R(q)=p*q=(180-(q)/(50))q=180q-(q^2)/(50) \\ \therefore R(q)=180q-(q^2)/(50) \end{gathered}`

C) We'll go ahead and determine the marginal revenue function, R'(q), by taking the derivative of the revenue function, R(q);

`\begin{gathered} \begin{equation*} R(q)=180q-(q^2)/(50) \end{equation*} \\ R^(\prime)(q)=180-(2q^(2-1))/(50)=180-(q)/(25) \\ \therefore R^(\prime)(q)=180-(q)/(25) \end{gathered}`

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;

`\begin{gathered} \begin{equation*} R^(\prime)(q)=180-(q)/(25) \end{equation*} \\ R^(\prime)(1100)=180-(1100)/(25)=180-44=136 \\ \therefore R^(\prime)(1100)=136 \end{gathered}`

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);

`\begin{gathered} P(q)=R(q)-C(q) \\ =(180q-(q^2)/(50))-(72,000+110q) \\ =180q-(q^2)/(50)-72000-110q \\ =-(q^2)/(50)+180q-110q-72000 \\ =-(q^2)/(50)+70q-72000 \\ \therefore P(q)=-(q^2)/(50)+70q-72000 \end{gathered}`

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);

`\begin{gathered} \begin{equation*} P(q)=-(q^2)/(50)+70q-72000 \end{equation*} \\ P^(\prime)(q)=-(2q^(2-1))/(50)+70(1*q^(1-1))-0=-2q^(50)+70q^0=-(q)/(25)+70 \\ \therefore P^(\prime)(q)=-(q)/(25)+70 \end{gathered}`

We can now go ahead and find P'(1100) as seen below;

`\begin{gathered} P^(\prime)(1100)=-(1100)/(25)+70=-44+70=26 \\ \therefore P^(\prime)(1100)=26 \end{gathered}`

So P'(1100) = 26