1. **State the problem:** We want to maximize the function $$f(q_1,q_2) = q_1^6 q_2^4 + 1.5 \ln q_1 + \ln q_2$$ subject to the constraint $$s_t = 100 = s_1 + 4 s_2$$. 2. **Set up the Lagrangian:** We introduce a Lagrange multiplier $\lambda$ for the constraint and write $$\mathcal{L} = q_1^6 q_2^4 + 1.5 \ln q_1 + \ln q_2 - \lambda (s_1 + 4 s_2 - 100)$$ 3. **Find partial derivatives:** $$\frac{\partial \mathcal{L}}{\partial q_1} = 6 q_1^5 q_2^4 + \frac{1.5}{q_1} - \lambda \frac{\partial s_1}{\partial q_1} = 0$$ $$\frac{\partial \mathcal{L}}{\partial q_2} = 4 q_1^6 q_2^3 + \frac{1}{q_2} - \lambda 4 \frac{\partial s_2}{\partial q_2} = 0$$ $$\frac{\partial \mathcal{L}}{\partial \lambda} = s_1 + 4 s_2 - 100 = 0$$ 4. **Assuming $s_1 = q_1$ and $s_2 = q_2$ (typical in such problems), the constraint is:** $$q_1 + 4 q_2 = 100$$ 5. **Rewrite the system:** $$6 q_1^5 q_2^4 + \frac{1.5}{q_1} = \lambda$$ $$4 q_1^6 q_2^3 + \frac{1}{q_2} = 4 \lambda$$ 6. **Divide the second equation by 4:** $$q_1^6 q_2^3 + \frac{1}{4 q_2} = \lambda$$ 7. **Set the two expressions for $\lambda$ equal:** $$6 q_1^5 q_2^4 + \frac{1.5}{q_1} = q_1^6 q_2^3 + \frac{1}{4 q_2}$$ 8. **Rearrange:** $$6 q_1^5 q_2^4 - q_1^6 q_2^3 = \frac{1}{4 q_2} - \frac{1.5}{q_1}$$ 9. **Factor left side:** $$q_1^5 q_2^3 (6 q_2 - q_1) = \frac{1}{4 q_2} - \frac{1.5}{q_1}$$ 10. **Use the constraint to express $q_1 = 100 - 4 q_2$ and substitute into the above equation to solve for $q_2$.** 11. **This is a nonlinear equation in $q_2$ that can be solved numerically.** 12. **Once $q_2$ is found, compute $q_1 = 100 - 4 q_2$.** 13. **Calculate $\lambda$ using either equation for verification.** **Final answer:** The optimal values $(q_1^*, q_2^*)$ satisfy $$q_1 + 4 q_2 = 100$$ and $$q_1^5 q_2^3 (6 q_2 - q_1) = \frac{1}{4 q_2} - \frac{1.5}{q_1}$$ which can be solved numerically for $q_2$ and then $q_1$.