By using slack(free) variable, solutions of both min and max problems

Question

$$F(x,y)= 2x^2-3y^2-2x$$. $$\text{s.t.} \ \ \ \ \ \ x^2+y^2\le 1$$

By using slack variable (free variable,), I want to solve minimum and maximum problems.

I want to solve this problem for maximum and minimum just like the solution of the above example. I dont want to solve this question in other ways. I cannot do it. Please help me thank you

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For example, $$\min f(x,y)= (x-a)^2 +b$$ $$\text{s.t.} \ \ \ \ \ \ x\le c$$ where a,b and c are constant.

Solution:

Let $\theta ^2$ be a slack (free) variable such that $\ \ \ \theta^2= x-c \ \ $ i.e.$\ \ \ (x-c)-\theta^2=0$

Then the Lagrange function is following

$$L(x,\lambda, \theta)= (x-a)^2 +b+\lambda ( (x-c)-\theta^2)=0$$

$$\partial L/\partial x= 2(x-a)+\lambda=0$$

$$\partial L/\partial \lambda= x-c-\theta^2 =0$$

$$\partial L/\partial \theta= -2\theta\lambda=0$$

Now let's take $-2\theta\lambda=0$

(I) $\lambda^*=0, \ \theta\not=0$ Condition

$\theta^2=a-c>0 \to a>c$

(II) $\lambda^*\not=0, \ \theta=0$ Condition

$x^*=c$ I.e. $c>a$ (III) $\lambda^*=0, \ \theta=0$ Condition

Answer 1

Your function $F$ is defined on a disc of radius one, which is a compact set. Therefore, the maximums and minimums are either on the boundary of the disc, or strictly inside the disc.

• Case 1: suppose they are on the boundary. In this case, rewrite your problem in polar coordinates as follows:

$$ F(r,t)=2r^2\cos^2(t)-3r^2\sin^2(t)-2r\cos(t), \quad \mbox{subject to } r\le 1. $$

But if we are one the boundary, then $r=1$ and your function $F$ is equal to $$F(t)=2\cos^2(t)-3\sin^2(t)-2\cos(t).$$

Derivative with respect to variable $t$ gives you the maximums and minimums on the boundary , namely:

$$ \max{F}=4\quad \mbox{for } (x,y)=(-1,0)\\ \min{F}=-\frac{16}{5}\quad \mbox{for } (x,y)=(\frac{1}{5},\frac{2\sqrt{6}}{5}) $$

• Case 2: suppose they are strictly inside the disc. In this case, the gradient of $F$ equals:

$$ \nabla F=(2x-2,-6y), $$

So potential maximums/minimums are have coordinates $(x,y)=(1,0)$ (which indeed is inside the disc). In this case,

$$ F(1,0)=0 $$

We can conclude from these case studies that the maximums and minimums are on the boundary: the maximum is $4$ and the minimum is $-16/5$.

Answer 2

Rewrite your function as $$2x^2-3y^2-2x = -3(x^2+y^2) + 5x^2-2x = -3(x^2+y^2) + 5\left(x^2-\frac 25x+\frac{1}{25}\right)-\frac{1}{5} = -3(x^2+y^2) + 5\left(x -\frac{1}{ 5}\right)^2-\frac{1}{5}.$$

Let the variable $r=\sqrt{x^2+y^2}$, then you obtain $$F=-3r^2 + 5\left(x -\frac{1}{ 5}\right)^2-\frac{1}{5}$$ with $r\in[0,1]$ and $x\in [-r,r]$.

If you want to find the minimum, then the solution is evident: we take the maximum possible value of $r$ (i.e. $r=1$) and $x=\frac 15$ (it falls in the range $[-1,1]$). The resulting minimum is $\inf F=-\frac{16}{5}$.

If you want to find the maximum, consider first the maximization problem for $5\left(x -\frac{1}{ 5}\right)^2$ on the interval $x\in[-r,r]$. Clearly, the maximum is attained when $|x-\frac 15|$ is maximized, hence we take $x=-r$. THerefore, we can say that

$$\sup F = \sup_{r\in[0,1]}\left( -3r^2 + 5\left(-r -\frac{1}{ 5}\right)^2-\frac 15\right) = \sup_{r\in[0,1]}\left( 2r^2 +2r \right) =4.$$