Suppose $x$ and $y$ are real numbers and $x^2+9y^2-4x+6y+4=0$. Then find the maximum value of $\left(\frac{4x-9y}{2}\right)$

Question

Suppose $x$ and $y$ are real numbers and $x^2+9y^2-4x+6y+4=0$. Then find the maximum value of $\left(\frac{4x-9y}{2}\right)$

Answer 1

Lagrange multipliers method

Maximize $$k \left(x^2-4 x+9 y^2+6 y+4\right)+\frac{1}{2} (4 x-9 y)$$

$ \left\{ \begin{array}{l} k (2 x-4)+2=0 \\ k (18 y+6)-\frac{9}{2}=0 \\ x^2-4 x+9 y^2+6 y+4=0 \\ \end{array} \right. $

$$\left(x=\frac{6}{5},\;y=-\frac{2}{15}\right);\;\left(x=\frac{14}{5},\;y=-\frac{8}{15}\right)$$

$$\frac{1}{2} (4 x-9 y)=8 \text{ is maximum for }x= \frac{14}{5},y= -\frac{8}{15}$$

$\color{red}{Second \;method}$

$r:\frac{1}{2} (4 x-9 y)=h$

$h$ is maximum (or minimum) when the line is tangent to the ellipse

$ \left\{ \begin{array}{l} \frac{1}{2} (4 x-9 y)=h \\ x^2-4 x+9 y^2+6 y+4=0 \\ \end{array} \right. $

$y =-\frac{2}{9} (h - 2 x)$

substitute

$$\frac{4}{9} (h-2 x)^2-\frac{4}{3} (h-2 x)+x^2-4 x+4=0$$ Expand and collect

$25x^2-4x(4 h+3) x+36-12h+4h^2=0$

to be tangent the discriminant must be zero

$\Delta=(4(4h+3))^2-100(36-12h+4h^2)=0$

$-144 \left(h^2-11 h+24\right)=0\to h_1=3;\;h_2=8$

so $\frac{1}{2} (4 x-9 y)=h$ is maximum when $h=8$

Answer 2

Use parameterisation,

$$x^2+9y^2-4x+6y+4=(x-2)^2+(3y+1)^2- 1 = 0$$

Let $x - 2 = \cos t$ and $3y + 1 = \sin t$, then

$$f(t) := \dfrac{4x - 9y}{2} = \dfrac{4(\cos t + 2) - 3(\sin t - 1)}{2} = \dfrac{4\cos t - 3\sin t + 11}{2}$$

Equating first derivative to zero we get, $f^\prime(t) = 0$ if $x = 2n \pi + 2 \tan^{-1}(3)$ or $x = 2n \pi - 2 \tan^{-1}\left(\dfrac 13\right)$ where $n \in \Bbb Z$.

Using second derivative test the maxima is given at $x = 2n \pi - 2 \tan^{-1}\left(\dfrac 13\right)$ and the maxima is $f\left(-\tan^{-1}\left(\dfrac13\right)\right) = 8$.