1. **Problem statement:** We have the function $$F(x_1, x_2) = 8x_1^{0.76}x_2^{0.12}$$ and the point $$a = (8, 3)$$. We want to find the exact change in $$x_2$$ when $$x_1$$ decreases by 0.25 units, keeping $$F$$ constant at $$F(a)$$. 2. **Formula and rule:** Since $$F$$ remains constant, we use the total differential set to zero: $$dF = \frac{\partial F}{\partial x_1} dx_1 + \frac{\partial F}{\partial x_2} dx_2 = 0$$ 3. **Calculate partial derivatives:** $$\frac{\partial F}{\partial x_1} = 8 \times 0.76 x_1^{0.76 - 1} x_2^{0.12} = 6.08 x_1^{-0.24} x_2^{0.12}$$ $$\frac{\partial F}{\partial x_2} = 8 \times 0.12 x_1^{0.76} x_2^{0.12 - 1} = 0.96 x_1^{0.76} x_2^{-0.88}$$ 4. **Evaluate partial derivatives at $$a = (8,3)$$:** $$\frac{\partial F}{\partial x_1}(8,3) = 6.08 \times 8^{-0.24} \times 3^{0.12}$$ Calculate powers: $$8^{-0.24} = \frac{1}{8^{0.24}} \approx \frac{1}{1.6596} = 0.6027$$ $$3^{0.12} \approx 1.1402$$ So, $$\frac{\partial F}{\partial x_1}(8,3) \approx 6.08 \times 0.6027 \times 1.1402 = 4.18$$ Similarly, $$\frac{\partial F}{\partial x_2}(8,3) = 0.96 \times 8^{0.76} \times 3^{-0.88}$$ Calculate powers: $$8^{0.76} \approx 5.18$$ $$3^{-0.88} = \frac{1}{3^{0.88}} \approx \frac{1}{2.63} = 0.3802$$ So, $$\frac{\partial F}{\partial x_2}(8,3) \approx 0.96 \times 5.18 \times 0.3802 = 1.89$$ 5. **Set total differential to zero and solve for $$dx_2$$:** $$dF = 0 = \frac{\partial F}{\partial x_1} dx_1 + \frac{\partial F}{\partial x_2} dx_2$$ $$\Rightarrow dx_2 = - \frac{\frac{\partial F}{\partial x_1}}{\frac{\partial F}{\partial x_2}} dx_1$$ Substitute values: $$dx_2 = - \frac{4.18}{1.89} \times (-0.25) = -2.21 \times (-0.25) = 0.55$$ 6. **Interpretation:** When $$x_1$$ decreases by 0.25, $$x_2$$ must increase by approximately 0.55 to keep $$F$$ constant. **Final answer:** $$\boxed{0.55}$$