By bounding the function, I need to find the maximum point of $7x+12y$ over the set $2x^2+6xy+9y^2-2x-6y\le 24$ (i.e., no Lagrange multipliers are allowed, etc.).
I am trying to show that $(7x+12y)^2\le C$ for a constant $C$. Then I will search for a point $(x_0,y_0)$ such that $7x_0+12y_0=\sqrt C$.
I did show that $(7x+12y)^2\le2944+18x^2+72xy$, but now I am stuck with $18x^2+72xy$.
Does anyone have some bright ideas? I am really stuck.
On
Inequality. Lagrange multiplier excluded.
$2x^2+6xy+9y^2-2x-6y\le 24$
$\implies x^2+(x+3y)-2x-6y^\le 24$
let $z=x+3y$, we have:
$x^2+z^2-2z\le24$
$\implies x^2+(z-1)^2\le25,$
which is a circle. And the function whose value is to maximized can be transformed to:
$7x+12y=7x+12\cdot\cfrac{z-x}{3}=3x+4z$,
let $c=3x+4z$, we have $z=\cfrac{-3x+c}{4}$, this is a line in xz plane with slope $-\cfrac 34$, and we want to maximize the y axis of the points of all lines with the given slope and intersect(or touch) to cirle. From the geometric interpretation of the problem, we know that the line pass the point on top-right of the circle with direvative of -0.75. the value of $c$ associate with it, the maximum value required, can be arrived at accordingly.
Hint: complete the squares in the inequality to get $x^2+(3y+x-1)^2\le 25$, then change $y$ to $z=3y+x-1$ to maximize $3x+4z$ subject to $x^2+z^2\le 25$. Now the method in the comment @jyre works. Otherwise a parameterization of the circle and maximizing $3\cos\theta+4\sin\theta$ works as well.