The weekly marginal cost functions C'(x) and the weekly marginal revenue function R'(x) of a company assembling personal computers are given by C'(x) =RM(3x^2-118x+1315) and R'(x)=RM(1000-4x), where x is the number of computers assembled per week. If the fixed costs are RM5000, find

Question

The weekly marginal cost functions C'(x) and the weekly marginal revenue function R'(x) of a company assembling personal computers are given by C'(x) =RM(3x^2-118x+1315) and R'(x)=RM(1000-4x), where x is the number of computers assembled per week. If the fixed costs are RM5000, find

a) the profit function
b) the increase in profit if the number of computers assembled increases from 30 to 35 per week
c) the maximum profit per week
d) the total revenue obtained and the price per unit when profit is maximised.
Please help. thank you

Answer 1

Since P(x) = R(x)-C(x)

P'(x) = R'(x)-C'(x)
max profit occurs where R'(x) = C'(x)

Since you have R' and C', you can see that

C(x) = x^3-59x^2+1315x + 5000
R(x) = 1000x-2x^2 + k
no idea what k is, since you don't specify what R is for any particular x. It is probably -5000, since there is that fixed cost, and zero production results in zero income.

Anyway, using those facts, you can answer all the questions.

Answer 2

To solve this problem, we'll perform the following steps:

Step 1: Integrate the marginal cost function, C'(x), to find the total cost function, C(x).
Step 2: Integrate the marginal revenue function, R'(x), to find the total revenue function, R(x).
Step 3: Subtract the total cost from the total revenue to determine the profit function, P(x).
Step 4: Use calculus to find the maximum profit, the quantity of computers that yields it, and the corresponding price per unit.

Let's go through each step in detail:

Step 1: Integrating C'(x) to find C(x):
C'(x) = RM(3x^2 - 118x + 1315)

To find C(x), we'll integrate C'(x) with respect to x:
C(x) = ∫ [RM(3x^2 - 118x + 1315)] dx
= RM ∫ (3x^2 - 118x + 1315) dx
= RM [x^3 - 59x^2 + 1315x] + C1

Here, C1 represents the constant of integration.

Step 2: Integrating R'(x) to find R(x):
R'(x) = RM(1000 - 4x)

To find R(x), we'll integrate R'(x) with respect to x:
R(x) = ∫ [RM(1000 - 4x)] dx
= RM ∫ (1000 - 4x) dx
= RM [1000x - 2x^2] + C2

Here, C2 represents the constant of integration.

Step 3: Determining the profit function, P(x):
To find the profit function, P(x), we'll subtract the total cost, C(x), from the total revenue, R(x):
P(x) = R(x) - C(x)
= (RM [1000x - 2x^2] + C2) - (RM [x^3 - 59x^2 + 1315x] + C1)
= RM (1000x - 2x^2) - RM (x^3 - 59x^2 + 1315x) + (C2 - C1)
= RM (1000x - 2x^2 - x^3 + 59x^2 - 1315x) + (C2 - C1)
= RM (-x^3 + 57x^2 + 685x) + (C2 - C1)

Step 4: Finding the maximum profit, quantity, and price per unit:
To find the maximum profit, we'll need to find the critical points of the profit function, P(x). We'll take the derivative of P(x) with respect to x, set it equal to zero, and solve for x.
P'(x) = -3x^2 + 114x + 685

Setting P'(x) = 0:
-3x^2 + 114x + 685 = 0

We can solve this quadratic equation to find the critical points.

Once we find the critical points, we can substitute these values into the profit function, P(x), to find the corresponding maximum profit and the quantity that yields it.

For part d, we can use the maximum profit value to determine the total revenue obtained (R) and the price per unit (P) when profit is maximized.
R = R(x)
P = R(x)/x

I hope this explanation helps you solve the problem! If you need any further assistance, feel free to ask.