1. **Problem Statement:** Given the total cost function $$TC = 100 - 12Q + 0.6Q^2$$, find the output level where marginal cost (MC) equals average variable cost (AVC). Also find total fixed cost (TFC), total variable cost (TVC), AVC, and minimum values. 2. **Formulas and Definitions:** - Marginal Cost (MC) is the derivative of total cost with respect to output $Q$: $$MC = \frac{dTC}{dQ}$$ - Total Fixed Cost (TFC) is the cost when output $Q=0$. - Total Variable Cost (TVC) is $TC - TFC$. - Average Variable Cost (AVC) is $$AVC = \frac{TVC}{Q}$$ - To find the output where $MC = AVC$, set the expressions equal and solve for $Q$. 3. **Calculate MC:** $$MC = \frac{d}{dQ}(100 - 12Q + 0.6Q^2) = 0 - 12 + 1.2Q = 1.2Q - 12$$ 4. **Calculate TFC:** At $Q=0$, $$TFC = TC(0) = 100 - 12\times0 + 0.6\times0^2 = 100$$ 5. **Calculate TVC:** $$TVC = TC - TFC = (100 - 12Q + 0.6Q^2) - 100 = -12Q + 0.6Q^2$$ 6. **Calculate AVC:** $$AVC = \frac{TVC}{Q} = \frac{-12Q + 0.6Q^2}{Q} = -12 + 0.6Q$$ 7. **Set MC equal to AVC and solve for $Q$:** $$1.2Q - 12 = -12 + 0.6Q$$ 8. **Simplify:** $$1.2Q - 12 = -12 + 0.6Q$$ $$1.2Q - 0.6Q = -12 + 12$$ $$0.6Q = 0$$ $$Q = 0$$ 9. **Interpretation:** The only solution is $Q=0$, which is trivial. To find the minimum AVC, differentiate AVC and set derivative to zero. 10. **Find minimum AVC:** $$AVC = -12 + 0.6Q$$ Derivative: $$\frac{dAVC}{dQ} = 0.6$$ Since derivative is positive constant, AVC is increasing with $Q$, so minimum AVC is at lowest positive $Q$ (approaching zero). 11. **Summary:** - $TFC = 100$ - $TVC = -12Q + 0.6Q^2$ - $MC = 1.2Q - 12$ - $AVC = -12 + 0.6Q$ - $MC = AVC$ only at $Q=0$ This means the marginal cost curve intersects the average variable cost curve at $Q=0$, and AVC increases linearly with $Q$. **Final answer:** - Marginal cost: $$MC = 1.2Q - 12$$ - Total fixed cost: $$TFC = 100$$ - Total variable cost: $$TVC = -12Q + 0.6Q^2$$ - Average variable cost: $$AVC = -12 + 0.6Q$$ - Output where $MC = AVC$: $$Q = 0$$