Find the minimum value of $9x^2+2y^2-8xy-6x+11$

Question

For any $x, y$ in real, find the minimum value of $9x^2+2y^2-8xy-6x+11$

Answer 1

Let $$9x^2+2y^2-8xy-6x+11=k$$ Thus, the the following quadratic equation of $y$ $$2y^2-8xy+9x^2-16x+11-k=0$$ has real solutions, which says $$16x^2-2(9x^2-6x+11-k)\geq0$$ or $$(x-3)^2\leq k-2,$$ which gives $$k\geq2.$$ The equality occurs for $x=3$ and $y=6,$ which says that $2$ a minimal value.

Answer 2

You can use some concepts from multivariable calculus, namely looking for where the gradient vector of the function is equal to $\langle 0,0\rangle$. (This does not return a minimum, but rather, critical points). $$\nabla f(x,y) = \langle18x -8y - 6, 4y - 8x\rangle$$ Solving for $18x - 8y - 6 =0$ and $4y -8x =0$, we find that a critical point exists at $(3,6)$. Plugging this in, we find that the minimum of the function is $f(3,6) = 2$.

Answer 3

Finding the critical point of the quadric by computing partial derivatives, as suggested in other answers, is pretty quick and easy, but you can always rewrite the quadric as a linear combination of squared terms by making the ansatz suggested here. In fact, there’s a purely mechanical algorithm akin to Gaussian elimination, which you can read about here.

If you start off with an integer matrix, this algorithm uses only integers, which is handy when you’re calculating by hand, but they can get quite large. I’m willing to work with rational numbers, so I’ll use an equivalent algorithm that uses the operation $R_i\to R_i-{a_{i1}\over a_{11}}R_1$ to clear the column below a pivot, and similarly for the column operations. Omitting the gory details, this algorithm starts with $$\left[\begin{array}{c|c}A&I\end{array}\right] = \left[\begin{array}{ccc|ccc}9&-4&-3 & 1&0&0 \\ -4&2&0 & 0&1&0 \\ -3&0&11 & 0&0&1\end{array}\right]$$ and produces $$\left[\begin{array}{c|c}D&P^T\end{array}\right]=\left[\begin{array}{ccc|ccc}9&0&0 & 1&0&0 \\ 0&\frac29&0 & \frac49&1&0 \\ 0&0&2 & 3&6&1 \end{array}\right].$$ We can already see from this that the minimum value is $2$, but continuing on, we have $D=P^TAP$, but want $A=P^{-T}DP^{-1}$, so we need to invert $P$. This can be done by applying the inverses of the column operations that were used to produce $P$, reversed, to the identity matrix. (This is just a variation of the well-known algorithm for inverting a matrix via Gaussian elimination.) This results in $$P^{-1} = \begin{bmatrix}1&-\frac49&-\frac13\\0&1&-6\\0&0&1\end{bmatrix},$$ so that $$9x^2-8xy+2y^2-6x+11 = 9\left(x-\frac49y-\frac13\right)^2+\frac29\left(y-6\right)^2+2(1)^2.$$

Answer 4

Any extreme values of $f(x,y)$ must occur at critical points, that is, at points where $f_x = f_y = 0$. In this example, taking these derivatives gives you a system of linear equations.

Answer 5

You would need to solve the system of linear equations given by partial derivates

$$\left\{ \begin{array}{l} \frac{\partial}{\partial x}f(x,y)=0\\ \frac{\partial}{\partial y}f(x,y)=0 \end{array}\right. \Rightarrow \left\{ \begin{array}{l} 18x-8y-6=0 \\ 4y-8x=0 \end{array}\right. \Rightarrow \left\{ \begin{array}{l} y=2x \\ y=6 \end{array}\right. $$

Answer 6

Define a function $F(x,y) = 9x^2 + 2y^2 - 8xy - 6x +11$

Then according to the Lagrange Multiplier Method, we know that there is no restriction, so we can define function $G(x,y)=F(x,y)+0\lambda=F(x,y)$,and then we take the partial derivatives of $x$ and $y$:

$$G'x(x, y) = 18x - 8y - 6 = 0$$

$$G'y(x, y) = 4y -8x = 0$$

Solve the equations, so we discover

$$x = 3, y = 6$$

Calculate it, it's the minimum value 2.

Answer 7

Among several methods, a general one to find the extremums of a function $f(x,y)$ consists in solving the system : $$\begin{cases} \frac{\partial f}{\partial x}=0 \\ \frac{\partial f}{\partial y}=0 \end{cases}$$ This leads to one or several points $(x_k,y_k)$ . For each one compute the corresponding value of $f(x_k,y_k)$. Comparing them (if they are several) allows to see if each point corresponds to a maximum or a minimum of $f(x,y)$ , local or general.

Several extremums can be observed for example in case of some polynomials of degree >2.

In the case of $$f(x,y)=9x^2+2y^2-8xy-6x+11$$ $$\begin{cases} \frac{\partial f}{\partial x}=18x-8y-6=0 \\ \frac{\partial f}{\partial y}=4y-8x=0 \end{cases}$$ Solving is very easy and gives : $x=3$ and $y=6$. $$f(3,6)=9(3^2)+2(6^2)-8(3)(6)-6(3)+11$$ $$f(3,6)=2$$ There is only one extremum. Another point (any one), for example $f(0,0)=11>f(3,6)$ . Thus the extremum is a minimum and the minimum value of the function is $2$.