1. The problem statement, all variables and given/known data a match box consists of an outer cover, open at both ends, into which slides a rectangular box without a top. The length of the box is one and a half times its breadth, the thickness of the material is negligible, and the volume of the box is 25cm^3 . If the breadth of the box is x cm, find , in terms of x, the area of the material used. Hence show that, if the least area of material is to be used to make the box , the length should be 3.7cm approximately. 2. Relevant equations 3. The attempt at a solution So I have this so far: 25 = 3/2 x^2 h [Volume and h is the height] For the surface area now, I figured I would find the surface area of the outer box(box1) and the inner box(box2) and then add them together. so I have this : box1 = ((3x^2)/2 + xh) (2) = 3x^2 + 2xh box 2 = (3x^2)/2 + xh(2) + (3xh/2)(2) = (3x^2)/2 + 5xh box1 + box2 = (9x^2)/2 + 7xh [ h = (50)/(3x^2) , from the volume ] Area(A) = (9x^2)/2 + 350/(3x) dA/dx = 9x - (350)/(3x^2) for max/min dA/dx = 0 9x = 350/(3x^2) x^3 = 350/27 x = 2.45cm (which is incorrect) Where did I go wrong?