
Larry C. answered 03/05/18
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Consider the border as made up of 4 rectangles. Along the narrow side, the two both have a length of 6 and width of b. Along the long side, each has a length of 10 + 2b and a width of b. So, the total maximum area of 80 is also:
2(6b) + 2(b(10+2b)) = 80 => 12b+20b+4b2=80 => 4b2+32b-80=0 which can then be solved via the quadratic formula