A long rectangular sheet of metal, 12 inches wide, is to be made into a rain gutter by turning up two sides so that they are perpendicular to the sheet. How many inches should be turned up to give the gutter its greatest capacity?

Answer 1

All we have to consider is the area of the cross-section, since the length is a constant.

let the sides to be turned up at both ends be x inches long
let the remaining base be y inches
2x + y = 12
y = 12-2x

area = xy = x(12-2x) = 12x - 2x^2
d(area)/dx = 12 - 4x = 0 for a max of area
4x = 12
x = 3

So the turn-ups should be 3 inches

Answer 2

Greatest capacity will be achieved when the area of cross section will be maximum.
Let X be the base ,so (12-X)/2 will be the height.
Area= Base x height= X(12-X)/2
=(12X-X^2)/2
Following the maximum theory, we differentiate the area w.r.t X and equate it to zero.
d/dX of(Area)=(12-2X)/2 =0
or 6-X=0
or X=6 inches(base)
so, height= (12-6)/2 = 3 inches

Answer 3

x (2x-2)
d/dX of(Area)=(12-2X)/2 =0
or 6-X=0
or X=6 inches(base)
so, height= (12-6)/2 = 3 inches

Answer 4

7inches long 5 inches wide