Given,
x−2y+4≥0,x+2y2≥0,x+4y2≤8,y≥0
Now, finding the point of intersection of x-2y+4=0&x+2{y}^{2}=0 we get,
(x,y)≡(−2,1)
And point of intersection of x-2y+4=0&x+4{y}^{2}=8 will be, (x,y)≡(−1,23)
Now, plotting the diagramw we get,

Now, from the above diagram, the required area will be,
A=∫−2−1(2x+4−2−x)+∫−10(28−x−2−x)dx+∫0828−xdx
⇒A=21[2(x+4)2]−2−1−21[(−x)23(3−2)]−2−1+21[(8−x)23(3−2)]−10−21[(−x)23(3−2)]−10+21[(8−x)23(3−2)]08
⇒A=9+45−34
⇒A=12107
⇒m+n=119