Vtyjkyuk

Consider two functions $f,g : X times Y subseteq mathbb R^2 to mathbb R$. Suppose that $f$ and $g$ are of class $mathcal C^2(X times Y)$. Suppose $(x^*,y^*)$ is the unique maximizer of our orginal program:
beginalign
&(x^*,y^*) = argmax_(x,y) in X times Yf(x,y) cdot g(x,y)\
texts.t. quad & f(x,y) geq 0\
& g(x,y) geq 0
endalign
I was wondering if we can reduce the dimensionality by first considering
beginalign
(x(alpha),y(alpha)) = argmax_(x,y) in X times Yleft[alpha f(x,y) + (1-alpha) g(x,y)right]
endalign
and then solve the parametrized program
beginalign
&alpha^* = argmax_alpha in [0,1]f(x(alpha),y(alpha)) cdot g(x(alpha),y(alpha))\
texts.t. quad & f(x(alpha),y(alpha)) geq 0\
& g(x(alpha),y(alpha)) geq 0
endalign

Question
Is the following generally true:
beginalign
(x^*,y^*) = (x(alpha^*),y(alpha^*))?
endalign

Example
Let $X = Y = [0,1]$ and $f(x,y) = x(1-y)$ and $g(x,y) = y(1-x)$. Since $f,g geq 0$ for all $x,y$ we don't need to consider the constraints.
Then $(x^*,y^*) = (1/2,1/2)$ solve $max_(x,y)f(x,y)cdot g(x,y)$. The parametrized maximzers are
beginalign
(x(alpha),y(alpha)) = (1-alpha,alpha) = argmax_(x,y)[alpha f(x,y) + (1-alpha)g(x,y)]
endalign
and
beginalign
alpha^* = 1/2 = argmax_alpha f(1-alpha,alpha)cdot g(1-alpha,alpha).
endalign
Such that $(x^*,y^*) = (x(alpha^*),y(alpha^*))$ is true here.