Posted by **MEG** on Monday, November 15, 2010 at 9:06pm.

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?

- calculus -
**Reiny**, Monday, November 15, 2010 at 9:44pm
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

- calculus -
**profkn**, Tuesday, November 16, 2010 at 1:23am
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

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