Крайні значення раціонального дробу

           We are looking for the extreme values of some rational fractions of form 

$$\frac{Ax^2+2Bx+C}{ ax^2+2bx+c}.$$

     We find in the book Daniel SITARU - Fenomen Algebric

     PROBLEM #1, page 27

In translation (thanks to Miss Google for this):

     " Find the minimum of the expression 

$$E=\frac{4ab-11a^2-14b^2}{3(a^2+b^2)}\; ;\;a,b \in \mathbb{R^*}."$$ 

 

ANSWER CiP

           We will find out more, both the minimum and the maximum of the expression.

$$-5=E(a,-2a)\leqslant  E(a,b) \leqslant E(2b,b)=-\frac{10}{3}\;\;,a,b \in \mathbb{R^*}.$$ 

 

Solution CiP

      The  number $\lambda$ is a value of the fraction $E$ if, and only if

" there are two numbers $a$ and $b$ such that $\lambda =E(a,b)"

$$\Leftrightarrow \; \exists \;a,b \in \mathbb{R^*}\;such\; that\;\lambda =\frac{4ab-11a^2-14b^2}{3(a^2+b^2)}$$

 $$\Leftrightarrow\;\exists\;a,b \in \mathbb{R^*}\;such\;that\;4ab-11a^2-14b^2=3\lambda a^2+3 \lambda b^2$$

 $$\Leftrightarrow\;\exists\;a,b \in \mathbb{R^*}\;such\;that\;(3\lambda+11)a^2-4ab+(3\lambda+14)b^2=0$$

$$\Leftrightarrow\; t=\frac{a}{b}\;is\;a\;root\;of\;the\;equation\;(3\lambda+11)t^2-4t+(3\lambda+14)=0 \tag{1}$$

$\Leftrightarrow\;\Delta _t\;\geqslant 0,\tag{2}$

where $\Delta _t$ is the half-discriminant to the equation (1),

$\Delta _t=4-(3\lambda+11)(3\lambda+14).$

But $(2)\;\Leftrightarrow\;9\lambda ^2+75 \lambda+150 \leqslant 0\;\Leftrightarrow\;3\lambda^2+25\lambda+50\leqslant 0\;\Leftrightarrow\;\lambda \in [-5;-\frac{10}{3}]$.

Here, the ends of the range $[-5;-\frac{10}{3}]$ are the roots of the equation $3\lambda^2+25\lambda+50=0.$

           Result that we have  $\lambda _{min}=-5,\;\lambda _{max}=-\frac{10}{3}.$

Further, 

$E(a,b)=-5\;\Leftrightarrow\;-4a^2-4ab-b^2=0\;\Leftrightarrow\;-(2a+b)^2=0\;\Leftrightarrow\;b=-2a$,

 and $E(a,b)=-\frac{10}{3}\;\Leftrightarrow\;a^2-4ab+4b^2=0\;\Leftrightarrow\;(a-2b)^2=0\;\Leftrightarrow\;a=2b.$

 We got the answer.

$\blacksquare$

 

Solution #2

(the authors' solution, only for the minimum, completed by CiP for the maximum)

 See page 72.

           Completed by CiP: $E=\frac{-10a^2-10b^2-a^2+4ab-4b^2}{3(a^2+b^2)}=-\frac{10}{3}-\frac{(a-2b)^2}{3(a^2+b^2)}\leqslant -\frac{10}{3}$,  $max\;E=-\frac{10}{3}$ for $a-2b=0.$

$\blacksquare$