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**Application of Derivatives** **Problem:** The cost in rupees of producing *x* items in a factory each day is given by $C(x) = 0.0013x^3 + 0.002x^2 + 5x + 2200$. Find the marginal cost when 150 items are produced. **Solution:** Let MC be the marginal cost. Total Cost $C(x) = 0.00013x^3 + 0.002...

**Application of Derivatives** **Problem:** The cost in rupees of producing *x* items in a factory each day is given by $C(x) = 0.0013x^3 + 0.002x^2 + 5x + 2200$. Find the marginal cost when 150 items are produced. **Solution:** Let MC be the marginal cost. Total Cost $C(x) = 0.00013x^3 + 0.002x^2 + 5x + 2200$ $MC = \frac{dC(x)}{dx}$ $MC = \frac{d}{dx}(0.00013x^3 + 0.002x^2 + 5x + 2200)$ $MC = 0.00039x^2 + 0.004x + 5$ To find the marginal cost when 150 items are produced, substitute $x = 150$ into the equation for MC: $MC = 0.00039(150)^2 + 0.004(150) + 5$ $MC = 0.00039(22500) + 0.6 + 5$ $MC = 8.775 + 0.6 + 5$ $MC = 14.375$ Therefore, the marginal cost when 150 items are produced is 14.375 rupees.