Answer:
(a)Revenue=580x-10x²,
Marginal Revenue=580-20x
(b)Fixed Cost, =900
Marginal Cost=300+50x
(c)Profit Function=280x-900-35x²
(d)x=4
Step-by-step explanation:
The price of a cake, p = 580 − 10x where x is the number of cakes sold per day.
Total cost function,c = (30+5x)²
(a)
<u>Revenue Function</u>
R(x)=x*p(x)=x(580 − 10x)
R(x)=580x-10x²
<u>Marginal Revenue Function</u>
This is the derivative of the revenue function.
If R(x)=580x-10x²,
R'(x)=580-20x
(b)Total cost function,c = (30+5x)²
c=(30+5x)(30+5x)
(x)=900+300x+25x²
Therefore, Fixed Cost=900
<u>Marginal Cost Function</u>
This is the derivative of the cost function.
If c(x)=900+300x+25x²
Marginal Cost, c'(x)=300+50x
(c)
<u>Profit Function</u>
Profit, P(x)=R(x)-C(x)
=(580x-10x²)-(900+300x+25x²)
=580x-10x²-900-300x-25x²
P(x)=280x-900-35x²
(d)To maximize profit, we take the derivative of P(x) in (c) above and solve for its critical point.
Since P(x)=280x-900-35x²
Derivative: P'(x)=280-70x
Equate to zero
280-70x=0
280=70x
x=4
The number of cakes which maximizes profit is 4.
