Exercise 1: average cost in dollars of manufacturing a certain item is given by: C(q) = 10 + 108/q + 1/2q^2 where q is the number (in thousands of units) of items produced. Find the minimum value of C and the corresponding total cost at that level of production.

To find the minimum value of the cost function C(q) and the corresponding total cost at that level of production, you need to take the derivative of the cost function with respect to q, set it equal to zero to find the critical points, and then determine whether each critical point is a minimum or maximum. Here's how you can do it step by step: Given: C(q) = 10 + 108/q + (1/2)q^2 Step 1: Take the derivative of C(q) with respect to q: C'(q) = d/dq [10 + 108/q + (1/2)q^2] C'(q) = -108/q^2 + q Step 2: Set the derivative equal to zero and solve for q to find the critical points: -108/q^2 + q = 0 Rearrange the equation: q^3 = 108 Take the cube root of both sides: q = 6 So, q = 6 is a critical point. Step 3: Determine if it's a minimum or maximum by using the second derivative test. Take the second derivative of C(q) with respect to q: C''(q) = d^2/dq^2 [-108/q^2 + q] C''(q) = 216/q^3 + 1 Now, evaluate C''(6): C''(6) = 216/(6^3) + 1 C''(6) = 216/216 + 1 C''(6) = 1 + 1 C''(6) = 2 Since the second derivative is positive at q = 6, it means that q = 6 corresponds to a local minimum. Step 4: Find the minimum value of C(q) and the corresponding total cost: Plug q = 6 back into the original cost function C(q): C(6) = 10 + 108/6 + (1/2)(6^2) C(6) = 10 + 18 + 18 C(6) = 46 So, the minimum value of C is 46 dollars, and the corresponding total cost at that level of production is $46.