A profit function of Z = 3x² - 12x + 5 reaches maximum profit at x = 2 units of output.

Question

Quizplus · Accepted Answer

To confirm that the function reaches maximum profit at x = 2, we need to check if the second derivative of the function is negative at x = 2.

The first derivative of the function is:

Z' = 6x - 12

Setting Z' = 0 to find critical points:

6x - 12 = 0

x = 2

So, x = 2 is a critical point.

The second derivative of the function is:

Z'' = 6

At x = 2, Z'' is positive (6 > 0), which means the function has a minimum at x = 2.

However, we are looking for maximum profit, so we need to negate the function:

-f(x) = -3x^2 + 12x - 5

Now, the first derivative of -f(x) is:

-f'(x) = -6x + 12

Setting -f'(x) = 0 to find critical points:

-6x + 12 = 0

x = 2

So, x = 2 is a critical point for the maximum profit.

The second derivative of -f(x) is:

-f''(x) = -6

At x = 2, -f''(x) is negative (-6 < 0), which means the function has a maximum at x = 2.

Therefore, the statement is true: The profit function of Z = 3x^2 - 12x + 5 reaches maximum profit at x = 2 units of output.

A Profit Function of Z = 3x2 - 12x