Respuesta :
Answer:
(a)Revenue function, [tex]R(x)=580x-x^2[/tex]
Marginal Revenue function, R'(x)=580-2x
(b)Fixed cost =900 .
Marginal Cost Function=300+50x
(c)Profit,[tex]P(x)=-35x^2+280x-900[/tex]
(d)x=4
Step-by-step explanation:
Part A
Price Function[tex]= 580 - 10x[/tex]
The revenue function
[tex]R(x)=x\cdot (580-10x)\\R(x)=580x-x^2[/tex]
The marginal revenue function
[tex]\dfrac{dR}{dx}= \dfrac{d}{dx}(R(x))=\dfrac{d}{dx}(580x-x^2)=580-2x\\R'(x)=580-2x[/tex]
Part B
(Fixed Cost)
The total cost function of the company is given by [tex]c=(30+5x)^2[/tex]
We expand the expression
[tex](30+5x)^2=(30+5x)(30+5x)=900+300x+25x^2[/tex]
Therefore, the fixed cost is 900 .
Marginal Cost Function
If [tex]c=900+300x+25x^2[/tex]
Marginal Cost Function, [tex]\frac{dc}{dx}= (900+300x+25x^2)'=300+50x[/tex]
Part C
Profit Function
Profit=Revenue -Total cost
[tex]580x-10x^2-(900+300x+25x^2)\\580x-10x^2-900-300x-25x^2\\$Profit,P(x)=-35x^2+280x-900[/tex]
Part D
To maximize profit, we find the derivative of the profit function, equate it to zero and solve for x.
[tex]P(x)=-35x^2+280x-900\\P'(x)=-70x+280\\-70x+280=0\\-70x=-280\\$Divide both sides by -70\\x=4[/tex]
The number of cakes that maximizes profit is 4.
