(10 points) A manufacturer of a particular product makes x bats at a cost of C(x)=8 x+20 dollars. The revenue from the sale of the product is given by R(x)=100 x-0.5 x^2 dollars.
i. (6 points) How many products should be manufactured and sold to maximize profit?
ii. (4 points) Use the first or second derivative to show that the number of products you found gives a maximum profit

Step 1 ：Given the cost function \(C(x) = 8x + 20\) and the revenue function \(R(x) = 100x - 0.5x^2\), we can find the profit function \(P(x)\) by subtracting the cost from the revenue: \(P(x) = R(x) - C(x) = -0.5x^2 + 92x - 20\).

Step 2 ：To find the number of products that maximize the profit, we need to find the derivative of the profit function \(P'(x)\) and set it equal to zero. This will give us the critical points. The derivative of the profit function is \(P'(x) = 92 - x\).

Step 3 ：Setting \(P'(x)\) equal to zero gives us the critical point \(x = 92\).

Step 4 ：We then use the second derivative test to determine whether this critical point is a maximum or minimum. The second derivative of the profit function is \(P''(x) = -1\).

Step 5 ：Since the second derivative at the critical point is negative, this means that the profit is maximized when 92 products are manufactured and sold.

Step 6 ：Substituting \(x = 92\) into the profit function gives us the maximum profit: \(P(92) = -0.5(92)^2 + 92(92) - 20 = 4212\).

Step 7 ：\(\boxed{\text{i. The manufacturer should manufacture and sell 92 products to maximize profit.}}\)

Step 8 ：\(\boxed{\text{ii. The second derivative of the profit function at x=92 is negative, which confirms that the profit is maximized at this point. The maximum profit is $4212.}}\)