The given revenue and cost functions are:
Revenue: R(x) = 50x - 0.6x^2
Cost: C(x) = 0.1x^2 + 15
The profit function is defined as the difference between the revenue and cost functions:
Profit: P(x) = R(x) - C(x) = 49x - 0.7x^2 - 15
To find the maximum profit, we need to find the maximum value of the profit function. We can do this by finding the critical points of the function and then applying the second derivative test.
The critical points of the profit function are the points where the derivative is equal to zero. The derivative of the profit function is:
P'(x) = 49 - 1.4x
Setting P'(x) equal to zero, we get:
49 - 1.4x = 0
1.4x = 49
x = 35/1.4 = 25
The critical point is x = 25. The second derivative of the profit function is:
P''(x) = -1.4
P''(25) = -1.4 < 0, which means that the profit function is decreasing at the critical point. Therefore, the maximum profit is achieved at x = 25.
The maximum profit is:
P(25) = 49(25) - 0.7(25)^2 - 15 = 415 - 350 - 15 = 40
Therefore, the maximum profit is $40, and the number of units that must be produced and sold in order to yield the maximum profit is 25.