1. **State the problem:** You have 3 lumber units to spend. Each lumber can buy 160000 HP directly. Each lumber can also buy 640 DEF, and each point of DEF increases your total HP by 2%. We want to find the best way to allocate lumber to maximize your total HP. 2. **Define variables:** Let $x$ be the number of lumber used to buy HP. Let $y$ be the number of lumber used to buy DEF. Since you have 3 lumber: $$x + y = 3$$ 3. **Express HP and DEF in terms of $x$ and $y$:** - HP bought directly: $$HP_{base} = 160000 \times x$$ - DEF bought: $$DEF = 640 \times y$$ 4. **Calculate effective HP including DEF buffs:** Each DEF point increases HP by 2%, so total multiplier from DEF is: $$M = 1 + 0.02 \times DEF = 1 + 0.02 \times 640y = 1 + 12.8y$$ 5. **Total effective HP:** $$HP_{total} = HP_{base} \times M = 160000x (1 + 12.8y)$$ 6. **Express $x$ in terms of $y$:** $$x = 3 - y$$ 7. **Rewrite total HP as function of $y$ only:** $$HP_{total}(y) = 160000(3 - y)(1 + 12.8y)$$ Expanding: $$HP_{total}(y) = 160000(3 + 38.4y - y - 12.8y^2) = 160000(3 + 37.4y - 12.8y^2)$$ $$HP_{total}(y) = 480000 + 5984000 y - 2048000 y^2$$ 8. **Find maximum by differentiating and setting derivative to zero:** $$\frac{d HP_{total}}{d y} = 5984000 - 4096000 y = 0$$ Solve for $y$: $$4096000 y = 5984000 \\ y = \frac{5984000}{4096000} = 1.46$$ 9. **Check if $y=1.46$ is within bounds $0 \leq y \leq 3$:** It is. 10. **Calculate $x$:** $$x = 3 - 1.46 = 1.54$$ 11. **Final allocation:** - Use approx 1.54 lumber to buy HP. - Use approx 1.46 lumber to buy DEF. 12. **Calculate final total HP:** $$HP_{base} = 160000 \times 1.54 = 246400$$ $$DEF = 640 \times 1.46 = 934.4$$ Multiplier: $$1 + 0.02 \times 934.4 = 1 + 18.688 = 19.688$$ Total HP: $$246400 \times 19.688 = 4,851,123.2$$ **Answer:** Allocate about 1.54 lumber to HP and 1.46 lumber to DEF to maximize your total effective HP of approximately 4,851,123.