Price of input, Px=13-2x-y; price of output, Py=13-x-2y; total cost, Tc=x+y.
Revenue comes from the sale of the output=y*Py. Variable cost comes from production of the input=x*Px.
Let the fixed cost be F, profit P=(revenue)-(total cost of production)=
yPy-xPx-F=13y-xy-2y^2-13x+2x^2+xy-F=
13(y-x)-2(y^2-x^2)-F=13(y-x)-2(y-x)(y+x)-F=(y-x)(13-2(y+x))-F,
and total cost is cost of production xPx+F=Tc=x+y (given). So 13x-2x^2-xy+F=x+y.
From this, 12x-2x^2+F=y(1+x) and y=(2x(6-x)+F)/(1+x). This relates output and input quantities.
So, dividing by 1+x: y=2(7-x)+(F-14)/(1+x). Since y>0, F>2x(x-6) and x>6 because F cannot be negative.
Px=13-2x-(2x(6-x)+F)/(1+x)=(13+13x-2x-2x^2-12x+2x^2-F)/(1+x)=(13-F-x)/(1+x). Px>0, so F<13-x, and x<13. Now we can see that 6<x<13. This implies that 1<F<6. But F>2x(x-6), so 0<F<182, assuming x is not confined to integers.
Py=13-x-2(2x(6-x)+F)/(1+x)=(13+13x-x-x^2-24x+4x^2-2F)/(1+x)=(13-2F-12x+3x^2)/(1+x)>0.
Therefore, 13-2F-12x+3x^2>0 because Py>0. x^2-4x+(13-2F)/4>0; x^2-4x+4+(13-2F)/4-4>0; (x-2)^2>(3+2F)/4. This implies x>2±sqrt(3/4+F/2)>0 and since we know 6<x<13, 16<(x-2)^2<121, so 30.5<F<241.5, so combining two conditions for F, we have 30.5<F<182. This contradicts the requirement 1<F<6. The full implication is that, we cannot find positive values for all of x, y, Px and Py, and the problem is unsolvable.