Answer
Using a contour map (please see the figure attached), we explain why $f\left( {x,y} \right)$ does not have a maximum subject to $g\left( {x,y} \right) = 0$ unless $g = af + b$ for some constants $a$, $b$.
Work Step by Step
Suppose that both $f\left( {x,y} \right)$ and the constraint function $g\left( {x,y} \right)$ are linear. So, $f\left( {x,y} \right)$ and $g\left( {x,y} \right)$ are represented by lines.
Suppose that $g = af + b$ for some constants $a,b \ne 0$. Taking the gradient of both sides gives
$\nabla g = a\nabla f$
So, $\nabla f = \frac{1}{a}\nabla g$.
By setting $\lambda = \frac{1}{a}$ and the constraint $g\left( {x,y} \right) = 0$, we obtain the Lagrange condition $\nabla f = \lambda \nabla g$ according to Theorem 1.
Since $\lambda$ is a scalar, this implies that the gradient $\nabla f$ is parallel to the gradient $\nabla g$.
But $\nabla g$ is perpendicular to the line $g\left( {x,y} \right)$ and $\nabla f$ is perpendicular to the line $f\left( {x,y} \right)$. Therefore, we conclude that the two lines $f\left( {x,y} \right)$ and $g\left( {x,y} \right)$ are parallel. By the method of Lagrange multipliers, the two lines coincide at critical point (the black line in the figure attached).
As is illustrated in the contour maps, there are level curves of $f$ that are smaller and larger values than the one that pass through the critical point, and that (the dashed lines in the figure attached) do not coincide with $g$. Thus, $f$ has a maximum subject to $g$ when $g = af + b$.
Hence, $f\left( {x,y} \right)$ does not have a maximum subject to $g\left( {x,y} \right) = 0$ unless $g = af + b$ for some constants $a$, $b$.
