Asked by MEG
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?
Answers
Answered by
Reiny
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
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
Answered by
profkn
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
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
Answered by
Anonymous
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
d/dX of(Area)=(12-2X)/2 =0
or 6-X=0
or X=6 inches(base)
so, height= (12-6)/2 = 3 inches
Answered by
zoe
7inches long 5 inches wide
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