Minimise $f:\mathbb{R}^2 \rightarrow \mathbb{R},(x,y) \mapsto f(x,y)=6x^2+3y^2-6xy+15x-9y+1$ under the condition that
(a) $2x+y\leq 4$
and separately under the condition
(b) $2x+y \geq4$
I'd normally use the method of Lagrange multipliers, but first: I think you'd have to be able to state the conditions in a way that $g(x,y)=0 $ and to me it isn't obvious how. Furthermore, it was given as a hint to just look for the usual minima and maxima of the function and then "check the rest by plugging it in".
So I found an extremum ( the only one ) at $(-1,\frac{1}{2})$ and because the Hessian is positiv definite ( I calculated the eigenvalues ) it's a minimum.
Now for $(-1,\frac{1}{2})$ (a) holds, but (b) doesn't. I can't see how both could hold for $2x+y \neq 4$ and since there is no minimum for which that is true, how do I minimise the function? Does there simply not exist a minimum?
Hence, we get a value $-\frac{35}{4}$.
We'll prove that it's a mimimal value.
Indeed, we need to prove that $$6x^2+3y^2-6xy+15x-9y+1+\frac{35}{4}\geq0$$ or $$2x^2-(2y-5)x+y^2-3y+\frac{13}{4}\geq0,$$ for which it's enough to prove that $$(2y-5)^2-8\left(y^2-3y+\frac{13}{4}\right)\leq0$$ or $$(2y-1)^2\geq0,$$ which is obvious.
By the same way we can prove that for $(x,y)=(0.65,2.7)$ we'll get a minimal value $0.325$.
Done!