Finding the minimum of $f(x)=3x^2+4y^2+4xy-11x-6y$ where $x$ and $y$ are reals.

Question

Question:

Find the minimum of $f(x, y)=3x^2+4y^2+4xy-11x-6y$ where $x$ and $y$ are reals and characterise the instances of equality.

My approach:

My first thought was to factorise this into a square. However, that is clearly not possible. AM-GM and Jenson's does not apply here due to the $-11x-6y$. Cauchy-Schwarz and other inequalities do not seem to apply either. According to dcode.fr, the minimum is $-\dfrac{41}{4}$ and equality is achieved when $x=2$ and $y=-\frac{1}{4}$, but I do not know how to prove this. Any help will be appreciated!

Answer 1

When the minimum of $f(x, y)$ is achieved, by definition, the derivative of $f(x, y)$ with respect to both $x$ and $y$ must be $0$ (when the other is held constant). Hence, it suffices to find the derivative of $f(x, y)$ with respect to both $x$ and $y$.

The derivative of $f(x, y)$ with respect to $x$ is $$\begin{align} \frac{\partial f}{\partial x} &=\frac{\partial}{\partial x}\left[3x^2+4y^2+4xy-11x-6y\right] \\ &=\frac{\partial}{\partial x}\left[3x^2+(4y-11)x+(4y^2-6y)\right] \\ &=6x+4y-11 & \text{since }y\text{ is held constant} \end{align}$$

Similarly, the derivative of $f(x, y)$ with respect to $y$ is $$\begin{align} \frac{\partial f}{\partial y} &=\frac{\partial}{\partial y}\left[3x^2+4y^2+4xy-11x-6y\right] \\ &=\frac{\partial}{\partial y}\left[4y^2+(4x-6)y+(3x^2-11x)\right] \\ &=8y+4x-6 & \text{since }x\text{ is held constant} \end{align}$$

Suppose that the minimum of $f(x, y)$ is achieved when $(x, y)=(x_0, y_0)$. Then, we will have

$$ \begin{cases} 6x_0+4y_0=11 \\ 4x_0+8y_0=6 \\ \end{cases} \\ \implies \begin{cases} x_0=2 \\ y_0=-\frac{1}{4} \\ \end{cases} $$

Now, we can substitute this back into $f(x, y)$ to get $$f(x_0, y_0)=-10\frac{1}{4}$$

Now, it suffices to prove that this is indeed the global minimum. Consider the double partial derivative of $f(x, y)$.

The double derivative of $f(x, y)$ with respect to $x$ is $$\begin{align} \frac{\partial^2 f}{\partial x^2} &=\frac{\partial}{\partial x}\left[6x+4y-11\right] \\ &=6 \end{align}$$

Similarly, the double derivative of $f(x, y)$ with respect to $y$ is $$\begin{align} \frac{\partial^2 f}{\partial y^2} &=\frac{\partial}{\partial y}\left[8y+4x-6\right] \\ &=8 \end{align}$$

Lastly, the double derivative of $f(x, y)$ with respect to $xy$ is $$\begin{align} \frac{\partial^2 f}{\partial x\partial y} &=\frac{\partial}{\partial x}\frac{\partial f}{\partial y} \\ &=\frac{\partial}{\partial x}\left(8y+4x-6\right) \\ &=4 \end{align}$$

Now, consider the Hessian matrix. Its determinant will be $6\times8-(4)^2=32>0$. Coupled with the fact that $\dfrac{\partial^2 f}{\partial x^2}>0$, we can conclude that $-10\frac{1}{4}$ is indeed the local minimum. However, since $f(x, y)$ is also strictly convex, $-10\frac{1}{4}$ is also the global minimum.

Hence, the minimum of $f(x, y)$ is $-10\frac{1}{4}$ with equality if and only if $x=2$ and $y=-\frac{1}{4}$.

Answer 2

$$f_x=6x+4y-11, \, f_y=8y+4x-6.$$ They are zero only at $x=2, $ $y=-1/4$, so its our candidate to extremum. Now $$f_{xx}=6,\,f_{xy}=4\,, f_{yy}=8, $$ so the jacobian is $48-16>0$, so our point is a global minimum.

Answer 3

By direct calculation, one has \begin{align*} f(x,y) & = 3x^2+4y^2+4xy-11x-6y \\ & = (2y+x)^2 +2x^2 - 11x - 6y \\ & = \left(2y+x - \dfrac{3}{2} \right)^2 +2x^2-8x - \dfrac{9}{4} \\ & = \left(2y+x - \dfrac{3}{2} \right)^2 +2\left(x-2 \right)^2 - \dfrac{41}{4} \end{align*}

For every $(x,y) \in \mathbb{R}^2$, one has directly $f(x,y) \geq -\dfrac{41}{4}$.

And moreover, one deduces that \begin{align*} f(x,y) = -\dfrac{41}{4} & \Longleftrightarrow \left(2y+x - \dfrac{3}{2} \right)^2 +2\left(x-2 \right)^2 - \dfrac{41}{4} = - \dfrac{41}{4} \\ & \Longleftrightarrow \left(2y+x - \dfrac{3}{2} \right)^2 +2\left(x-2 \right)^2 =0 \end{align*}

so $f(x,y) = -\dfrac{41}{4} \Longleftrightarrow \left\lbrace \begin{array}{l} 2y+x - \dfrac{3}{2}=0 \\ x-2 = 0 \end{array} \right. \Longleftrightarrow \left\lbrace \begin{array}{l} x=2 \\ y = -\dfrac{1}{4}\end{array} \right.$

Answer 4

Solving $$ \frac{\partial f}{\partial x} = 6x + 4x -11 = 0 \\ \frac{\partial f}{\partial y} = 4x + 8y -6 = 0 \\ $$ shows that $(x_0, y_0) = (2, -1/4)$ is a stationary point and a possible candidate for the minimum. Now $$ f(u + 2 , v - \frac 14) = 3 u^2 + 4 v^2+4 uv- \frac{41}{4} = 2 u^2 + (u+2v)^2 - \frac{41}{4} $$ shows that $f(x, y) \ge - \frac{41}{4}$ with equality if and only if $u=v=0$, so that $(2, -1/4)$ is the global minimum.

Remark: It is not a coincidence that $f(x, y)$ can be written as a “sum of squares” plus the minimal value. This is generally true for quadratic functions of the form $$ f(x,y)= Ax^2+2Bxy +Cy^2 +2Dx + 2Ey + F $$ with $A > 0$ and $B^2 < AC$, see for example Finding minimum value of a function of two variables or Help in justifying that $f(a,b)$ is a global minimum.

One can also argue that $f$ is strictly convex because the Hessian matrix, in this case $$ H_f = \begin{pmatrix} \frac{\partial^2 f}{\partial x^2} && \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial x \partial y} && \frac{\partial^2 f}{\partial x^2} \\ \end{pmatrix} = \begin{pmatrix} 6 & 4 \\ 4 & 8 \end{pmatrix} $$ is positive definite everywhere. (Strictly) convex functions have the property that any stationary point is a (strict) global minimum.

Answer 5

Another way.

Let $x=2a$ and $y=-\frac{1}{4}b$.

Thus, $$3x^2+4y^2+4xy-11x-6y=12a^2+\frac{1}{4}b^2-2ab-22a+\frac{3}{2}b=$$ $$=\frac{1}{4}(48a^2-8ab+b^2-88a+6b)=\frac{1}{4}\left(\frac{1}{3}(12a-b-11)^2-\frac{(b+11)^2}{3}+b^2+6b\right)=$$ $$=\frac{1}{12}\left((12a-b-11)^2+2b^2-4b-121\right)=$$ $$=\frac{1}{12}\left((12a-b-11)^2+2(b-1)^2-123\right)\geq-\frac{123}{12}=-\frac{41}{4}.$$ The equality occurs for $a=b=1,$ which says that we got a minimal value.

Answer 6

In this answer, we apply the $\color{#c00}{\Delta}$ method, which is one of the general high school level methods that can work quickly for our purposes .

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Let

$$\begin{align}3x^2\!+\!4y^2\!&+\!4xy\!-\!11x\!-\!6y\!=\!a\end{align}$$

Let's rearrange the equation as a quadratic equation with respect to $\color{#c00}{x}$ :

$$\begin{align}3\color{#c00}{x}^2\!+\!\color{#c00}{x}(4y\!-\!11)\!&+\!(4y^2\!-\!6y\!-\!a)\!=\!0\end{align}$$

Then, we determine the discriminant $\Delta_{\color{#c00}{x}}$ :

$$ \begin{align}\Delta_{\color{#c00}{x}}&=-32\color{#0a0}{y}^2-16\color{#0a0}{y}+(121+12a)\\ &=-\left(32\color{#0a0}{y}^2+16\color{#0a0}{y}-(121+12a)\right)\\ &≥0\end{align} $$

We recall that, if $\Delta_{\color{#0a0}{y}}<0$, then $\Delta_{\color{#c00}{x}}<0$ . A contradiction . Therefore, we have $\Delta_{\color{#0a0}{y}}≥0$. Thus,

$$ \begin{align}&\Delta_{\color{#0a0}{y}}=64+32(121+12 a)≥0\\ \implies &a≥-\frac {41}{4}\thinspace . \end{align} $$

which completes the answer .