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

(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

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

`(dR)/(dx)= (d)/(dx)(R(x))=(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,
`(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.