Question

QUESTION 3 [10 MARKS] 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] (3 marks) B. Find the fixed cost and marginal cost function [Hint: fixed cost does not change with quantity produced] (3 marks) C. Find the profit function [Hint: profit is revenue minus total cost] (2 marks) D. Find the quantity that maximizes profit (2 marks)

Answer 1

Answer:

(a)Revenue function, $R(x)=580x-x^2$

Marginal Revenue function, R'(x)=580-2x

(b)Fixed cost =900 .

Marginal Cost Function=300+50x

(c)Profit,$P(x)=-35x^2+280x-900$

(d)x=4

Step-by-step explanation:

Part A

Price Function$= 580 - 10x$

The revenue function

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

The marginal revenue function

$\dfrac{dR}{dx}= \dfrac{d}{dx}(R(x))=\dfrac{d}{dx}(580x-x^2)=580-2x\\R'(x)=580-2x$

Part B

(Fixed Cost)

The total cost function of the company is given by $c=(30+5x)^2$

We expand the expression

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

Therefore, the fixed cost is 900 .

Marginal Cost Function

If  $c=900+300x+25x^2$

Marginal Cost Function, $\frac{dc}{dx}= (900+300x+25x^2)'=300+50x$

Part C

Profit Function

Profit=Revenue -Total cost

$580x-10x^2-(900+300x+25x^2)\\580x-10x^2-900-300x-25x^2\\ Profit,P(x)=-35x^2+280x-900$

Part D

To maximize profit, we find the derivative of the profit function, equate it to zero and solve for x.

$P(x)=-35x^2+280x-900\\P'(x)=-70x+280\\-70x+280=0\\-70x=-280\\ Divide both sides by -70\\x=4$

The number of cakes that maximizes profit is 4.