The Joys Of Minimalism

Relevant wiki: Hessian Matrix

The function is : $\large f_a(x,y)=ax^2+4ay^2+20xy+3x+6y+7$

Finding the first partial derivatives gives us : $\begin{cases} 2ax+20y+3=0 \\ 8ay+20x+6 = 0 \end{cases}$

Taking the Hessian we see,

$\Delta^2f_a(x,y) = \begin{pmatrix} 2a & 20 \\ 20 & 8a \end{pmatrix}$

For attaining a global minima it must be Positive Semi-Definite which implies in turn $f$ should be convex throughout the domain.

The principal minors must be non-negative , $16a^2-400\ge 0$ & $2a,8a\ge 0$ . We get on solving $a\ge 5$

$\begin{cases} \text{For a=5, } f_{5}(x,y)=(x+2y)(5x+10y+3)+7 \\ \text{For a>5, }f_{a}(x,y)\text{ is continously increasing for all positive a} \end{cases}$

There are critical points in this case as the coefficient vector of the linear terms $[3,6]^T$ is in the range of the Hessian. So $a=5$ gives local as well as global minimum for the function as the function $f$ is convex & it's Hessian is positive semi-definite.

We have: $\begin{aligned} f_a(x,y)&=5x^2+20xy+20y^2+3(x+2y)+7+(a-5)(x^2+4y^2)\\ &=5(x+2y)^2+3(x+2y)+7+(a-5)(x^2+4y^2)\\ &=5\left(x+2y+\dfrac{3}{10}\right)^2+\dfrac{131}{20}+(a-5)(x^2+4y^2) \end{aligned}$

If $a<5$ , we can choose $y=k; x=-2k-\dfrac{3}{10}$ and $k\to+\infty$ .

This implies $f_a(x,y)=(a-5)\left(\left(2k+\dfrac{3}{10}\right)^2+4k^2\right)\to +\infty$ when $k\to+\infty$

If $a=5$ , we get $f_a(x,y)=5\left(x+2y+\dfrac{3}{10}\right)^2+\dfrac{131}{20}\ge\dfrac{131}{20}$

The equality holds when $x+2y+\dfrac{3}{10}=0$ .

So $\boxed{a=5}$ is the smallest value satisfied.

We first prove that $a> 0$ : Suppose otherwise, that is suppose there exists a minimum value when $a\leq 0$ , then setting $y=0$ shows that $f_a(x,0) = ax^2 + 3x+7$ is either a straight line or a parabola that concave downwards. Either way, neither them has a minimum value. Thus, $a> 0$ must be true.

We can rewrite the matrix of the quadratic part of $f_a(x,y)$ as

$A_a = \begin{bmatrix}{a} && {10} \\ {10} && {4a}\end{bmatrix}$

By Cayley-Hamilton theorem , the eigenvalues of $A_a$ satisfy the equation $| \lambda I_2 - A | = 0$ , or equivalently,

$0 = \begin{vmatrix}{\lambda - a} && {-10} \\ {-10} && {\lambda - 4a}\end{vmatrix} \quad \Leftrightarrow \quad 0= (\lambda - a)(\lambda - 4a)-100 \quad \Leftrightarrow \quad 0 = \lambda^2 + \lambda(-5a) + (4a^2-100) \; .$

Applying the quadratic formula , we can solve for $\lambda$ to get $\lambda = \dfrac{5a\pm \sqrt{9a^2+400}}2$ .

The necessary requirement for the function $f_a(x,y)$ to have a minimum value, is to have all its eigenvalues to be strictly non-negative.

Let $\lambda _1 < \lambda_2$ denote these two eigenvalues. Because we have shown that $a> 0$ , then $\lambda_2\geq 0$ . We are left to determine the range of $a$ such that $\lambda_1 \geq 0$ as well. So,

$\dfrac{5a - \sqrt{9a^2+400}}2 \geq 0 \quad \Leftrightarrow \quad (25a)^2 \geq 9a^2 + 400 \quad \Leftrightarrow \quad a \geq 5 \; .$

Thus, we know that there is a minimum value when $a>5$ . What's left to prove is to determine whether $a=5$ gives a minimum $f_a(x,y)$ :

$f_5(x,y) = 5x^2 + 20y^2 + 20xy + 3x+6y + 7 \; .$

We then find the kernel of this quadratic form:

$\left [ \begin{array} { c c | c} 5 & 10 & 0 \\ 10 & 20 & 0 \end{array}\right ] \stackrel{R_2\to R_2 - 2R_1}{\huge \longrightarrow}\left [ \begin{array} { c c | c} 5 & 10 & 0 \\ 0 & 0 & 0 \end{array}\right ] \quad \Leftrightarrow \quad (x,y) = (-2t,t), t \in \mathbb R \; .$

Now $f_5(-2t,t) = 7$ , showing that $f_5(x,y)$ has a minimum value. And we're done! We successfully proved that $f_a(x,y)$ has a minimum value when $a\geq \boxed5$ .

The 2nd order equation

$a{x}^{2}+4a{y}^{2}+20xy+3x+6y+7=k$

becomes a family of hyperbolas for $\left| a \right| <5$ , because then

${B}^{2}-4AC={20}^{2}-4\cdot 1a\cdot 4a=400-16{a}^{2}>0$

Hence, we first seek if there is a value for $a\ge 5$ such that this reduces to a pair of lines. To find that, letting the equation equal $k$ , we solve for $y$

$y=\dfrac { 1 }{ 8a } \left( -6-20x\pm \sqrt { (\left( 400-16{ a }^{ 2 } \right) { x }^{ 2 }+\left( 240-48a \right) x+16ak-112a+36 } \right)$

We see that if $a=5$ , then the radical becomes independent of $x$ , and the problem is answered. For $a<-5$ , the 2nd order equation becomes a family of ellipses with a global maximum and no global minimum.

For the function (which is a second degree equation in 2 variables) to have a global minimum and a to be smallest , I think it must represent a parabola. So, H^2= A.B where 2H is the coefficient of xy , B is the coefficent of y^2 and A is the coefficient of x^2 . Therefore, a comes out to be 5.

Please improve my solution, I know I've answered it vaguely.

Despite not knowing about Hessian matrices, here is my solution. Taking partials, 2ax+20y=-3, 8ay+20x=-6. Now substitute y in terms of x and a and one gets 2a^2x-15-50x+3a=0. Tracking the discrimant for a to real in terms of x yields eventually x=-3/20. When we track this discriminant, we are finding the smallest x such that there is a corresponding real a, as this will occur in an increasing multivariable function. Plug this back into the equation 2ax+20y=-3, with y substituted, yielding a^2-10a+25=0. And the solution is obvious.

I DONE IT IN A DIFFERENT WAY

CLEARLY THE EQUATION IS HOMOGENOUS OF 2 DEGREE FOR MIN (global min) ITS DETERMINANT MUST BE ZERO (pair of straight lines) solving determinant will give directly 5