10. The cost, in thousand of dollars, of making $x$ items, in hundreds, is given by $C(x)=-12 x-75$.
The revenue function is given by $R(x)=-6 x^{2}+30$.
Determine the number of items that should be sold to maximize profit. (3 marks)

Step 1 ：The cost function is given by \(C(x)=-12x-75\) and the revenue function is given by \(R(x)=-6x^{2}+30\).

Step 2 ：The profit function is the difference between the revenue function and the cost function, which is \(P(x) = R(x) - C(x) = -6x^{2} + 12x + 105\).

Step 3 ：To find the maximum profit, we need to find the derivative of the profit function and set it equal to zero. The derivative of the profit function is \(P'(x) = 12 - 12x\).

Step 4 ：Setting the derivative equal to zero gives us the critical points. Solving \(P'(x) = 0\) gives us the critical point \(x = 1\).

Step 5 ：We need to verify if this point is a maximum or a minimum. We can do this by taking the second derivative of the profit function and substituting the critical point into it. The second derivative of the profit function is \(P''(x) = -12\).

Step 6 ：Substituting the critical point into the second derivative gives us \(P''(1) = -12\). Since this is negative, the critical point is a maximum.

Step 7 ：Therefore, the number of items that should be sold to maximize profit is the critical point, which is \(x = 1\). However, since x was in hundreds of items, we multiply the critical point by 100 to get the actual number of items.

Step 8 ：Final Answer: The number of items that should be sold to maximize profit is \(\boxed{100}\).