$(x-3)^2+(y-2)^2 \leq 20$ $(x-3+2y-4)^2 \leq ((x-3)^2+(y-2)^2)(1+4)=100 \to |x+2y-7| \leq 10 \to -3 \leq x+2y \leq 17$

One can also think geometrically: The region for $x,y$ is a disc around $(3,2)$ with radius $\sqrt{20}$ and we want to find the largest value $c$ such that the line $x+2y=c$ intersects this disc. But this will be the tangent line with slope $-\frac{1}{2}$ to the circle. We can find the touching point by taking the perpendicular diameter of the circle with slope $2$ and intersect: But this will just be the point $(5,6)$. Hence $x+2y=17$ is maximal.

Let $ x,y $ be reals such that $ x(x-6)\leq y(4-y)+7.$ Prove that$$ -2\leq 2x +y \leq 18$$$$-10\sqrt{2}-3 \leq x-3y \leq 10\sqrt{2}-3$$$$ 13-2\sqrt{65}\leq 3x +2y \leq 13+2\sqrt{65}$$

Let $ x,y $ be reals such that $2x +y=18.$ Prove that$$ x(x-6)+ y(y-4)\geq 7$$Let $ x,y $ be reals such that $2x +y=-2.$ Prove that$$ x(x-6)+ y(y-4)\geq 7$$ VicKmath7 wrote: The reals $x, y$ satisfy $x(x-6)\leq y(4-y)+7$. Find the minimal and maximal values of the expression $x+2y$.

sqing wrote: Let $ x,y $ be reals such that $ x(x-6)\leq y(4-y)+7.$ Prove that$$ -2\leq 2x +y \leq 18$$$$-10\sqrt{2}-3 \leq x-3y \leq 10\sqrt{2}-3$$$$ 13-2\sqrt{65}\leq 3x +2y \leq 13+2\sqrt{65}$$ More generally, $\forall x,y,\lambda\in\mathbb{R}$, if $ x(x-6)\leq y(4-y)+7$, then $$3+2\lambda-2\sqrt{5\left(1+\lambda^2\right)}\le x+\lambda y \leq 3+2\lambda+2\sqrt{5\left(1+\lambda^2\right)}.$$Solution. The assumption gives $$(x-3)^2+(y-2)^2\le20.$$By Cauchy-Schwarz’s inequality we get \begin{align*}&|x+\lambda y-(3+2\lambda)|\\ =&|(x-3)+\lambda(y-2)|\\ \leq &\sqrt{\left(1+\lambda^2\right)\left[(x-3)^2+(y-2)^2\right]}\\ \le&2\sqrt{5\left(1+\lambda^2\right)}, \end{align*}from which the conclusion follows. Taking $$\lambda\in\left\{\frac{1}{2},-3,\frac{2}{3}\right\}$$generates the 3 inequalities quoted. $\blacksquare$

sqing wrote: Let $ x,y $ be reals such that $2x +y=18.$ Prove that$$ x(x-6)+ y(y-4)\geq 7$$Let $ x,y $ be reals such that $2x +y=-2.$ Prove that$$ x(x-6)+ y(y-4)\geq 7$$ Solution. Using the two assumptions $$2x+y=18\text{ or }-2$$and by Cauchy-Schwarz’s inequality we get \begin{align*}&x(x-6)+ y(y-4)\\ =&(x-3)^2+(y-2)^2-13\\ =&\frac{2^2+1^2}{5}\left[(x-3)^2+(y-2)^2\right]-13\\ \ge &\frac{1}{5}\left[2(x-3)+(y-2)\right]^2-13\\ =&\frac{1}{5}\left[2x+y-8\right]^2-13=20-13=7.\,\, \blacksquare \end{align*}