Answer to Question #113888 in Calculus for ashmita

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

A.

Revenue function = (price per unit) $\times$ (quantity of units)

$$R=p(x)\cdot x=(580-10x)x=580x-10x^2$$

Total point

$$C(x)=(30+5x)^2=900+300x+25x^2$$

Marginal revenue

$$R'(x)={dR \over dx}=(580x-10x^2)'=580-20x$$

B.

$$C(x)=900+300x+25x^2$$

$$Fixed\ cost=900$$

$$Marginal\ cost=C'(x)=(900+300x+25x^2)'=$$ $$=300+50x$$

C.

Profit function = revenue − cost

$$P(X)=R(x)-C(x)=$$ $$=580x-10x^2-(900+300x+25x^2)=$$ $$=-35x^2+280x-900, x\geq0$$

Marginal profit

$$Marginal\ profit=P'(x)=$$ $$=(-35x^2+280x-900)'=-70x+280$$

$$P'(x)=0=>-70x+280=0=>x=4$$

If $0\leq x<4,$ then $P'(x)>0, P(x)$ increases.

If $x>4,$ then $P'(x)<0,P(x)$ decreases.

$$P(4)=-35(4)^2+280(4)-900=-340$$

The function $P(x)$ is the local maximum with value of $-340$ at $x=4.$

Since the function $P(x)$ has the only extremum, then the profit function has the absolute maximum with value of $-340$ at $x=4.$