State the Karush
-
Kuhn
-
Tucker Conditions
(
KKT
)

explicitly and solve or interpret them to determine candidate solutions.
The goal is to minimize the cost function:
Cost
=
f
(
x
)
=
4
3
0
0
x

Question

State the Karush
-
Kuhn
-
Tucker Conditions 
(
KKT
)
 
﻿
explicitly and solve or interpret them to determine candidate solutions.
The goal is to minimize the cost function:
Cost
=
f
(
x
)
=
4
3
0
0
x
^
2
+
3
3
0
0
y
^
2
+
1
3
0
0
x
-
4
2
0
y
+
1
6
8
0
0
Subject to the following constraints:
0
.
5
x
+
0
.
8
y 
<
=
 
4
2
X
^
2
+
Y
^
2

≥
 
2
0
X
^
2
 
≤
 
1
2
Y
^
2
 
≤
 
2
6

give the calculation step by step as like as full answer in 2 steps and explanation block in every steps. 
And final answer summary
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VIPSolutions · Accepted Answer

**Step 1: State the KKT Conditions and Problem Parameters**

---

**Definition of Parameters:**

Let  
$ x $ and $ y $ = decision variables.  
Cost function:  
$ f(x, y) = 4300x^2 + 3300y^2 + 1300x - 420y + 16800 $

Subject to constraints:  
1. $ 0.5x + 0.8y \leq 4 $  
2. $ x^2 + y^2 \geq 20 $  
3. $ x^2 \leq 12 $  
4. $ y^2 \leq 26 $

**KKT Conditions:**

For a minimization problem with constraints $ g_i(x) \leq 0 $ and $ h_j(x) = 0 $, the KKT conditions are:

1. **Stationarity:**  
   $ \nabla f(x^*) + \sum_{i=1}^{m} \lambda_i \nabla g_i(x^*) = 0 $
2. **Primal feasibility:**  
   $ g_i(x^*) \leq 0 $
3. **Dual feasibility:**  
   $ \lambda_i \geq 0 $
4. **Complementary slackness:**  
   $ \lambda_i g_i(x^*) = 0 $

**Constraints in Standard Form:**

1. $ g_1(x, y) = 0.5x + 0.8y - 4 \leq 0 $
2. $ g_2(x, y) = 20 - x^2 - y^2 \leq 0 $   (reverse sign for $ \geq $)
3. $ g_3(x, y) = x^2 - 12 \leq 0 $
4. $ g_4(x, y) = y^2 - 26 \leq 0 $

**Lagrangian:**

Let $ \lambda_1, \lambda_2, \lambda_3, \lambda_4 \geq 0 $ be Lagrange multipliers for each constraint.

$$
\mathcal{L}(x, y, \lambda) = 4300x^2 + 3300y^2 + 1300x - 420y + 16800
+ \lambda_1(0.5x + 0.8y - 4)
+ \lambda_2(20 - x^2 - y^2)
+ \lambda_3(x^2 - 12)
+ \lambda_4(y^2 - 26)
$$

---

**Explanation Block:**

In this step, all parameters and constraints are explicitly defined. The cost function and constraints are written in the standard form required for KKT. The KKT conditions are stated, and the Lagrangian is constructed using these constraints and their multipliers.

---

**Step 2: Solve for Stationarity and Candidate Solutions**

---

**Find gradients (set to zero for stationarity):**

$$
\frac{\partial \mathcal{L}}{\partial x} = 8600x + 1300 + 0.5\lambda_1 - 2x\lambda_2 + 2x\lambda_3 = 0
$$
$$
\frac{\partial \mathcal{L}}{\partial y} = 6600y - 420 + 0.8\lambda_1 - 2y\lambda_2 + 2y\lambda_4 = 0
$$

**Primal feasibility:**

Each constraint must be satisfied.

**Dual feasibility:**

Each $ \lambda_i \geq 0 $.

**Complementary slackness:**

For each $ i $, $ \lambda_i g_i(x^*, y^*) = 0 $.

---

**Case Study:**

First, check the unconstrained minimum (all $ \lambda_i = 0 $):

$$
8600x + 1300 = 0 \implies x = -\frac{1300}{8600} \approx -0.1512
$$
$$
6600y - 420 = 0 \implies y = \frac{420}{6600} \approx 0.0636
$$

Check if this point is feasible:

- $ 0.5x + 0.8y = 0.5(-0.1512) + 0.8(0.0636) = -0.0756 + 0.0509 = -0.0247 < 4 $ ✔️
- $ x^2 + y^2 = (-0.1512)^2 + (0.0636)^2 = 0.0229 + 0.0040 = 0.0269 < 20 $ ❌ (violates constraint 2)
- $ x^2 = 0.0229 < 12 $ ✔️
- $ y^2 = 0.0040 < 26 $ ✔️

Constraint $ x^2 + y^2 \geq 20 $ is violated, so the unconstrained minimum is infeasible.

---

**Now, enforce $ x^2 + y^2 = 20 $ (i.e., constraint 2 active, so $ \lambda_2 \geq 0 $)**

Let $ x^2 + y^2 = 20 $.  
Suppose other constraints are inactive ($ \lambda_1 = \lambda_3 = \lambda_4 = 0 $).

Stationarity:

$$
8600x + 1300 - 2x\lambda_2 = 0
$$
$$
6600y - 420 - 2y\lambda_2 = 0
$$

Solve for $ x $ and $ y $:

$$
x(8600 - 2\lambda_2) = -1300 \implies x = \frac{-1300}{8600 - 2\lambda_2}
$$
$$
y(6600 - 2\lambda_2) = 420 \implies y = \frac{420}{6600 - 2\lambda_2}
$$

But $ x^2 + y^2 = 20 $:

$$
\left( \frac{-1300}{8600 - 2\lambda_2} \right)^2 + \left( \frac{420}{6600 - 2\lambda_2} \right)^2 = 20
$$

Let $ \mu = \lambda_2 $. This is a nonlinear equation in $ \mu $. The solution involves solving for $ \mu $, then obtaining $ x $ and $ y $.

---

**Explanation Block:**

The gradients of the Lagrangian are used to find stationary points. The unconstrained minimum is found to be infeasible due to the circle constraint, so the circle constraint is enforced as equality (active). The problem reduces to solving a nonlinear equation for the Lagrange multiplier, then obtaining $ x $ and $ y $ from their expressions in terms of $ \mu $.

---

**Final Answer Summary**

- **Parameters:** All variables and constraints are defined, with constraints rewritten in KKT-friendly form.
- **KKT Setup:** The KKT conditions are explicitly stated, and the Lagrangian is constructed.
- **Stationarity:** The gradient equations are set up, and an initial (unconstrained) minimum is tested for feasibility and rejected.
- **Active Constraint:** The quadratic constraint ($ x^2 + y^2 \geq 20 $) must be active at the solution; thus, a system of equations is set up to solve for the candidate minimizer.
- **Candidate Solution:**  
  The optimal $ (x^*, y^*) $ on the circle $ x^2 + y^2 = 20 $ is determined by solving:

$$
  x = \frac{-1300}{8600 - 2\mu}
  $$
  $$
  y = \frac{420}{6600 - 2\mu}
  $$
  with
  $$
  x^2 + y^2 = 20
  $$
  where $ \mu \geq 0 $.

This is a nonlinear equation for $ \mu $, which can be solved numerically to find the candidate point. The other constraints should be checked for activity; if any are active, repeat the process with those constraints enforced as equalities.

**Interpretation:**  
The KKT condition

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