The total cost in dollars to produce $q$ units of a product is $C(q)$. Fixed costs are $\$ 22,500$. The marginal cost is \[ C^{\prime}(q)=0.008 q^{2}-q+54 \] Round your answers to two decimal places. (a) Find $C$ (150), the total cost to produce 150 units. The total cost to produce 150 units is $\$$ (b) Find the value of $C^{\prime}(150)$. \[ C^{\prime}(150)=\$ \mathbf{i} \] (c) Use parts (a) and (b) to estimate $C$ (151). \[ C(151)=\$ \mathbf{i} \] eTextbook and Media

Step 1 ：Find the antiderivative of \(C'(q)\) which is the integral of \(C'(q)\).

Step 2 ：The antiderivative of \(C'(q)\) is: \(\int C'(q) dq = \int (0.008q^2 - q + 54) dq = 0.008/3 * q^3 - 1/2 * q^2 + 54q + K\) where \(K\) is the constant of integration.

Step 3 ：We know that the fixed costs are $22,500, so \(K = 22,500\).

Step 4 ：So, \(C(q) = 0.008/3 * q^3 - 1/2 * q^2 + 54q + 22,500\).

Step 5 ：Substitute \(q = 150\) into \(C(q)\) to get \(C(150) = 0.008/3 * 150^3 - 1/2 * 150^2 + 54 * 150 + 22,500 = \$ 30,000\).

Step 6 ：\(\boxed{C(150) = \$ 30,000}\)

Step 7 ：To find \(C'(150)\), substitute \(q = 150\) into \(C'(q)\): \(C'(150) = 0.008 * 150^2 - 150 + 54 = \$ 120\).

Step 8 ：\(\boxed{C'(150) = \$ 120}\)

Step 9 ：To estimate \(C(151)\), use the formula \(C(151) \approx C(150) + C'(150)\): \(C(151) \approx 30,000 + 120 = \$ 30,120\).

Step 10 ：\(\boxed{C(151) \approx \$ 30,120}\)