Can somebody help me answer this marginal cost problem? I'm having a hard time doing so. Also, any tips

A company manufactures and sells x light bulbs per week. if the weekly cost is C=5,000 +2x and the weekly price per light bulb is p= 10-0.001x, 0 ≤ x ≤ 10,000 respectively, how many light bulbs should be manufactured and sold every week in order to maximize profit?


Please do help :) I'm having a hard time in my calculus class. Thank you, Good day! :)

Profit is maximum when the marginal revenue is equal to marginal cost.
Total revenue = R(x) = x*p = x(10-0.001x) = 10x-0.001x^2
Marginal revenue = d/dx(Total revenue)
= d/dx(10x-0.001x^2)
= d/dx(10x)-0.001*d/dx(x^2)
= 10-0.002x
Total cost = C(x) = 5000+2x
Marginal cost = d/dx(Total cost)
= d/dx(5000+2x)
= 0+2
= 2
Therefore 10-0.002x = 2
-0.002x = -8
x = 8/0.002 = 4000
4000 light bulbs should be manufactured and sold every week to maximize profit.
>>>>>>>>>>>>>>>>>.
The question can also be answered by maximizing the profit function
Profit = Total revenue-Total cost
P(x) = R(x)-C(x) = (10x-0.001x^2) - (5000+2x)
= 10x-0.001x^2-5000-2x
= -0.001x^2+8x-5000
For maximization of profit
first derivative of profit function P(x)
dP/dx = d/dx(-0.001x^2+8x-5000)
= -0.001*d/dx(x^2)+8*d/dx(x)-d/dx(5000)
= -0002x+8-0
= -0.002x+8
Set dP/dx = 0
-0.002x+8 = 0
-0.002x = -8
x = 8/0.002 = 4000
Second derivative of the profit function
d^2P/dx^2 = d/dx(dP/dx) = d/dx(-0.002x+8)
= -0.002* d/dx(x)+d/dx(8)
= -0.002+0
= -0.002
Since the second derivative is negative (d^P/dx^2 < 0), it can be said that the profit function has a maximum at x=4000
Maximum profit = P(x=4000) = -0.001*4000^2+8*4000-5000
= -16000+32000-5000
= 11000

To maximize profit; marginal cost = marginal revenue; i.e, MC=MR
The revenue = px
MC = d(C)/dx & MR = d(px)/dx
Therefore, d[5000+2x]/dx = d[(10-0.001x)x]/dx
2 = 10-0.002x
x= 4000 bulbs

I hope you understood