Question

Maximixing Profit. In Business, Profit is the difference between revenue and cost; that is, Total Profit= Total Revenue - Total cost, P(x)= R(x) - C(x), where x is the number of units sold. Find the maximum profit and the number of units that must be sold in order to yield the maximum profit for each of the folllowing. R(x) = 5x, C(x) = 0.001x^2 + 1.2x + 60

Answer 1

do you know about derivatives. (calculus)?

Answer 2

In calculus, a branch of mathematics, the derivative is a measure of how a function changes as its input changes

Answer 3

if you do, then

P(x) = 5x-( 0.001x^2 + 1.2x + 60)
P(x) = 5x-.001x^2-1.2x-60
P(x) = -.001x^2+3.8x-60

// this is the derivative of P(x)
P'(x) = -.002x+3.8

// set the derivative function P'(x) to zero to find the maximum value for the profit)
-.002x+3.8 = 0
-.002x = -3.8
x = 1900 units to be sold for maximum profit.

Answer 4

You can also check your answer from a graph:

https://www.google.com/search?q=5x-(+0.001x%5E2+%2B+1.2x+%2B+60)&oq=5x-(+0.001x%5E2+%2B+1.2x+%2B+60)&aqs=chrome.0.57&sourceid=chrome&ie=UTF-8

The max profit will be at the turning point... which appears to be x = 1900,

Answer 5

x is the number of units sold, so 1900 is the amount of units which results in the maximum amount of profit. Therefore, we can use the profit equation P(x) = -.001x^2+3.8x-60

Answer 6

(my computer printed it before i finished so just copied it again)
x is the number of units sold, so 1900 is the amount of units which results in the maximum amount of profit. Therefore, we can use the profit equation P(x) = -.001x^2+3.8x-60.


P(x) = -.001(1900)^2+3.8(1900)-60 = 3550

Answer 7

Thank you I was so confused.

Answer 8

yw