1. The problem is to minimize the function $$f(x_1,x_2) = -100 - x_2 + 0.01x_1^2 - 0.01x_1 + 10$$ subject to the constraints $$-15 \leq x_1 \leq 5$$ and $$-3 \leq x_2 \leq 3$$. 2. The function is quadratic in $x_1$ and linear in $x_2$. To find the minimum, we consider the partial derivatives and the constraints. 3. Compute the partial derivatives: $$\frac{\partial f}{\partial x_1} = 0.02x_1 - 0.01$$ $$\frac{\partial f}{\partial x_2} = -1$$ 4. Set the partial derivatives to zero to find critical points: $$0.02x_1 - 0.01 = 0 \implies x_1 = \frac{0.01}{0.02} = 0.5$$ Since $$\frac{\partial f}{\partial x_2} = -1$$ is constant and negative, $f$ decreases as $x_2$ increases. 5. Considering constraints: - For $x_1$, the critical point $0.5$ lies within $$[-15,5]$$. - For $x_2$, since $f$ decreases as $x_2$ increases, the minimum occurs at the maximum allowed $x_2 = 3$. 6. Evaluate $f$ at $$x_1=0.5$$ and $$x_2=3$$: $$f(0.5,3) = -100 - 3 + 0.01(0.5)^2 - 0.01(0.5) + 10$$ $$= -103 + 0.01 \times 0.25 - 0.005 + 10$$ $$= -103 + 0.0025 - 0.005 + 10$$ $$= -103 + 0.0025 - 0.005 + 10 = -92.9975$$ 7. Check boundaries for $x_1$ to ensure minimum: - At $x_1 = -15$, $x_2=3$: $$f(-15,3) = -100 - 3 + 0.01(225) - 0.01(-15) + 10 = -103 + 2.25 + 0.15 + 10 = -90.6$$ - At $x_1 = 5$, $x_2=3$: $$f(5,3) = -100 - 3 + 0.01(25) - 0.01(5) + 10 = -103 + 0.25 - 0.05 + 10 = -92.8$$ 8. The minimum value is at $$x_1=0.5$$ and $$x_2=3$$ with $$f = -92.9975$$. Final answer: $$\boxed{x_1 = 0.5, x_2 = 3, f_{min} = -92.9975}$$