2021/22 AHSMC Part 1 Problem 8

If a vertical cut is made in the block, then the sum of the surface areas of the two pieces is the surface area of the initial rectangular block plus twice the area of the vertical cut. The area of a vertical cut, which is a rectangle, is maximum if its base is the diagonal of a face of the rectangular box. The area of such a vertical cut can be $$3\sqrt{4^2 + 5^2} = 3\sqrt{41},\qquad 4\sqrt{3^2 + 5^2} = 4\sqrt{34},\qquad \text{or}\qquad 5\sqrt{3^2 + 4^2} = 25.$$ Since $$3\sqrt{41} < 4\sqrt{34} < 25,$$ the largest possible value for the sum of the surface area of the two pieces is $$2 \cdot 3 \cdot 4 + 2 \cdot 3 \cdot 5 + 2 \cdot 4 \cdot 5 + 2 \cdot 25 = 144.$$ The answer is (E).