Optimisation with differentiation

The total cost for the production and marketing of x items of a certain product is $C where
C = 5,000 + 60x
The revenue received from each item is $R where R = 300 - x
Given that : Profit = Revenue - Cost find the value of x that gives maximum profit and determine what this maximum profit will be.

My attempt:
P = (300 - x) - ( 5,000 + 60x)
P = 300 - x - 5,000 - 60x
P= 300 - 61x - 5,000
P = -61x -4,700
Differentiating to find the max point does not work here.


Can someone show me their working out?Thank you.

Answer:When x = 120 the maximum profit of $9400 is realised.

your original equation is flawed: instead of (300 - x), you should have x(300 - x) - since 300 - x is the revenue per unit, you must multiply it by the number of units marketed (x) to find the total revenue. The total cost is already adjusted for the amount, so it is unnecessary to alter this

thus: P = x(300 - x) - (5000 + 60x)
P = 300x - x^2 - 5000 - 60x
P = -x^2 + 240x - 5000

dP/dx = -2x + 240
let dP/dx = 0 to find the minimum:

0 = -2x + 240
2x = 240
x = 120

Sub this back in:

P = 120(300 - 120) - (5000 + 60*120)
P = 36000 - 14400 - 5000 - 7200
P = 36000 - (26600)
P = $9400

as required
Hope this helps.

Naw, I just left it somewhat unfinished because I assumed you could find P. :)

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Notice something. The cost is given as a total cost. Revenue is given per x, not as a total. So you need to multiply R by x to get the total revenue.

P = Rx - C
P = x(300 - x) - 5,000 - 60x
P = 300x - x² - 5,000 - 60x
P = -x² + 240x - 5000
P' = -2x + 240 = 0
2x = 240
x = 120

...which is consistent with the answer provided. Pay close attention to details.