Let's solve the problem step by step. --- ### **Given Problem:** \[ \text{Minimize: } 2x^2 - 18x + 2xy + y^2 - 16y + 56 \] \[ \text{Subject to: } x + 4y \leq 8 \] --- ## **(a) Find the minimum solution to this problem.** We have a quadratic objective function with a linear constraint. ### **Step 1: Formulate the Lagrangian** Let \(\lambda\) be the Lagrange multiplier for the constraint \(x + 4y \leq 8\): \[ L(x, y, \lambda) = 2x^2 - 18x + 2xy + y^2 - 16y + 56 + \lambda(x + 4y - 8) \] ### **Step 2: Take partial derivatives and set to zero** Find stationary points by setting the gradients to zero: \[ \frac{\partial L}{\partial x} = 4x - 18 + 2y + \lambda = 0 \] \[ \frac{\partial L}{\partial y} = 2x + 2y - 16 + 4\lambda = 0 \] \[ \frac{\partial L}{\partial \lambda} = x + 4y - 8 \leq 0, \quad \lambda \geq 0, \quad \lambda(x + 4y - 8) = 0 \] ### **Step 3: Solve the unconstrained minimum** First, try the unconstrained minimum (set \(\lambda = 0\)): \[ 4x - 18 + 2y = 0 \implies 4x + 2y = 18 \tag{1} \] \[ 2x + 2y - 16 = 0 \implies x + y = 8 \tag{2} \] From (2): \(x + y = 8 \implies y = 8 - x\) Substitute into (1): \[ 4x + 2(8-x) = 18 \implies 4x + 16 - 2x = 18 \implies 2x = 2 \implies x = 1 \] \[ y = 8 - 1 = 7 \] Check if this point satisfies the constraint: \[ x + 4y = 1 + 4 \times 7 = 1 + 28 = 29 > 8 \] So it **violates the constraint**. ### **Step 4: The constraint is binding** So the minimum will be on the boundary \(x + 4y = 8\). Let’s substitute \(x = 8 - 4y\) into the objective function and minimize with respect to \(y\): \[ f(x, y) = 2x^2 - 18x + 2xy + y^2 - 16y + 56 \] Plugging \(x = 8 - 4y\): \[ f(y) = 2(8-4y)^2 - 18(8-4y) + 2(8-4y)y + y^2 - 16y + 56 \] Let’s expand step by step: \[ (8-4y)^2 = 64 - 64y + 16y^2 \] So, \[ 2(8-4y)^2 = 2 \times (64 - 64y + 16y^2) = 128 - 128y + 32y^2 \] \[ -18(8-4y) = -144 + 72y \] \[ 2(8-4y)y = 16y - 8y^2 \] \[ y^2 = y^2 \] \[ -16y \] \[ +56 \] Sum all terms: \[ f(y) = [128 - 128y + 32y^2] + [-144 + 72y] + [16y - 8y^2] + y^2 - 16y + 56 \] Combine like terms: - Constants: \(128 - 144 + 56 = 40\) - \(y\) terms: \(-128y + 72y + 16y - 16y = -56y\) - \(y^2\) terms: \(32y^2 - 8y^2 + y^2 = 25y^2\) So: \[ f(y) = 40 - 56y + 25y^2 \] Now, minimize this quadratic function. The minimum is at: \[ y^* = -\frac{b}{2a} = -\frac{-56}{2 \times 25} = \frac{56}{50} = 1.12 \] Now, \(x^* = 8 - 4y^* = 8 - 4 \times 1.12 = 8 - 4.48 = 3.52\) Plug these into the constraint to check: \[ x^* + 4y^* = 3.52 + 4 \times 1.12 = 3.52 + 4.48 = 8 \quad \text{(satisfies constraint)} \] Now, compute the minimum value: \[ f(y^*) = 40 - 56 \times 1.12 + 25 \times (1.12)^2 = 40 - 62.72 + 31.36 = 8.64 \] So: \[ \boxed{8.64 \text{ at } (x, y) = (3.52, 1.12)} \] --- ## **(b) If the right-hand side of the constraint increases from 8 to 9, how much do you expect the objective function to change?** This is the **dual value (shadow price)** at the constraint. Recall that when the constraint is binding, the dual variable \(\lambda\) equals the rate of change of the optimal value with respect to the constraint. We had the Lagrangian stationarity conditions: \[ 4x - 18 + 2y + \lambda = 0 \] \[ 2x + 2y - 16 + 4\lambda = 0 \] Plug in \(x^*=3.52, y^*=1.12\): First equation: \[ 4(3.52) - 18 + 2(1.12) + \lambda = 0 \implies 14.08 - 18 + 2.24 + \lambda = 0 \implies -1.68 + \lambda = 0 \implies \lambda = 1.68 \] So, the objective function is expected to **decrease by 1.68** for an increase of 1 in the right-hand side. \[ \boxed{1.68} \] --- ## **(c) Resolve the problem with a new right-hand side of the constraint of 9.** Now, constraint: \(x + 4y = 9\) Let \(x = 9 - 4y\), plug into the objective function: \[ f(y) = 2(9-4y)^2 - 18(9-4y) + 2(9-4y)y + y^2 - 16y + 56 \] First, expand \((9-4y)^2 = 81 - 72y + 16y^2\): \[ 2(81 - 72y + 16y^2) = 162 - 144y + 32y^2 \] \[ -18(9-4y) = -162 + 72y \] \[ 2(9-4y)y = 18y - 8y^2 \] \[ y^2 \] \[ -16y \] \[ +56 \] Sum all: \[ f(y) = [162 - 144y + 32y^2] + [-162 + 72y] + [18y - 8y^2] + y^2 - 16y + 56 \] Constants: \(162 - 162 + 56 = 56\) \(y\) terms: \(-144y + 72y + 18y - 16y = -70y\) \(y^2\) terms: \(32y^2 - 8y^2 + y^2 = 25y^2\) \[ f(y) = 56 - 70y + 25y^2 \] Minimizing: \[ y^* = \frac{70}{50} = 1.4 \] \[ x^* = 9 - 4 \times 1.4 = 9 - 5.6 = 3.4 \] Check constraint: \(x^* + 4y^* = 3.4 + 5.6 = 9\) Compute the minimum value: \[ f(y^*) = 56 - 70 \times 1.4 + 25 \times (1.4)^2 = 56 - 98 + 49 = 7 \] So the new minimum value is \(\boxed{7}\). **Actual decrease:** \(8.64 - 7 = 1.64\) This is slightly less than the estimated \(1.68\) from the dual value. --- ### **Final Answers (for filling in the blanks):** **(a)** \[ \boxed{8.64} \text{ at } (x, y) = (\boxed{3.52},\,\boxed{1.12}) \] **(b)** \[ \boxed{1.68} \] **(c)** \[ \boxed{7},\ \boxed{1.64} \] --- **Summary Table:** | Part | Answer | |---|---| | (a) | 8.64 at (3.52, 1.12) | | (b) | 1.68 | | (c) | 7, 1.64 |