I am not sure what the second-order Lagrange condition is and how it applies to this?
Minimize $x^2 + y^2$
Subject to $x^2 - y - 4 \leq 0$ and $y - x - 2 \leq 0$.
Please can someone assist me in the steps/process to solve this using the Second-order Lagrange condition??
You just have to apply Lagrange Multiplier for inequality constrain, i.e. KKT conditions.
So we have two inequality constrains and we have to look into $2^2=4$ distinct cases, and the initial function:
$$f(x,y) = x^2 + y^2$$
becomes:
$$F(x,y,\lambda,\lambda_1) = x^2 + y^2 + \lambda(x^2 - y - 4) - \lambda_1(y-x-2)$$
Now we take partial derivatives and set them to $0$:
$$F_x = 2x + 2\lambda x + \lambda_1 = 0$$ $$F_y = 2y - \lambda - \lambda_1 = 0$$ $$F_{\lambda} = x^2 - y - 4 = 0$$ $$F_{\lambda_1} = y-x-2 = 0$$
Now we check all the cases:
Case 1: $\lambda = \lambda_1 = 0$
If we substitute back into the parital derivatives we get that $2x = 0 \implies x = 0$ and $2y = 0 \implies y = 0$. So we obtain one solution:
$$(x,y) = (0,0)$$
We can chech that that solution fulfil the constrains.
Case 2: $\lambda = y-x-2 = 0$
If we set $F_x = F_y$ we get: $y - x = \lambda_1 \implies \lambda_1 = 2 $. Now substituting back we have: $y= 1$ and $x=-1$. So we obtained another solution:
$$(x,y) = (-1,1)$$
Case 3: $x^2-y-4 = \lambda_1=0$
$F_x$ can be rearranged as $2x(1+\lambda) = 0$, becasue, the gradient $\lambda$ must be non-negative so we have $x=0$. From the condition we have that $y = -4$. So we have another solution:
$$(x,y) = (0,-4)$$
Case 4: $x^2-y-4 = y-x-2=0$
From the condition we can get: $y = x + 2$ and plugging into the first equation we have:
$$x^2 - y - 4 = 0$$ $$x^2 - x - 2 - 4 = 0$$ $$x^2 - x - 6 = 0$$
Solving this quadratic equation we have: $x_1 = 3$ and $x_2 = -2$. For $y$ we have $y=5$ and $y = 0$ respectively.
So we obtain two solutions:
$$(x,y) = (3,5), (-2,0)$$
Plugging those five solutions into the initial function, we can check that the minima occurs at $f(0,0) = 0$ and the maxima occurs at $f(3,5) = 9 + 25 = 34$