Answer to Question #113773 in Calculus for Archana

Question

A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the number of cakes sold per day, and p is price. The total cost function of the company is given by c = (30+5x) 2 where x is previously defined, and c is total cost.
A. Find the revenue and marginal revenue functions [Hint: revenue is price multiplied by quantity i.e. revenue = price × quantity]
B. Find the fixed cost and marginal cost function [Hint: fixed cost does not change with quantity produced]
C. Find the profit function [Hint: profit is revenue minus total cost]
D. Find the quantity that maximizes profit

Answer 1

Given, price function $p(x)=580-10x$ and total cost $C(x) = (30+5x)^2=25x^2+300x+900$.

A) Revenue function $R(x) = xp(x) = x(580-10x) = 580x-10x^2$

Marginal Revenue = $\dfrac{dR}{dx}=580-20x$

B) Total cost = Variable cost + Fixed cost $=25x^2+300x+900$

Therefore, fixed cost = 900.

Marginal cost = $\dfrac{dC}{dx}=50x-300$.

C) Profit function

$P(x)= R(x) - C(x) = 580x-10x^{2}-25x^{2}-300x-900\\ P(x) = -35x^{2}+280x-900$

D) Now, $P'(x)=0$, gives

$-70x+280 = 0 \Rightarrow x = 4$.

Since $\dfrac{d^{2}P}{dx^{2}}=-70<0, x=4$ maximizes the profit.