A piece of sheet metal is rectangular, 5ft wide 8ft long. Congurent squares are to be cut from its 4 corners.?

The resulting piece of metal is to be folded and welded to form an open-topped box. How should this be done to get a box of the largest possible volume?A piece of sheet metal is rectangular, 5ft wide 8ft long. Congurent squares are to be cut from its 4 corners.?
Let four a ft x a ft squares be cut from the 4 corners


Dimensions of the open top box are (8 - 2a) ft x (5 - 2a) ft x a ft


We must have 0 %26lt; 2a %26lt; 5 (thus 0 %26lt; 2a %26lt; 8 also) (i. e.) 0 %26lt; a %26lt; 5/2


Measure of its volume V = 4a^3 -26 a^2 + 40a


dV / da = 12 a^2 - 52 a + 40 = 4 ( 3 a^2 - 13 a + 10) = 4 ( a - 1) ( 3a - 10)


a %26lt; 5/2 implies we cannot have 3a - 10 =0


So dV/da vanishes when a = 1.


Now d^2y / da^2 = 24 a - 52 which is negative at a = 1


Hence V is maximum when a = 1 and


max V = 6 x 3 x 1 = 18 c. ft.

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