Question

guys please help: A channel is to be constructed out of a sheet of metal, to allow gravel to be removed from a mining site. The sheet of metal has a width 40cm, and each side is to be bent up (at the same angle theta=pi/4 from the horizontal) so as to maximise the amount of gravel that the channel can carry-see the following sketch (in comments below).

Answer 1

Let L be the length of the base.
Width of sheet metal =
2 * w * cos(45) + L = 2 * w * sqrt(2) / 2 + L =
w * sqrt(2) + L = 40
L = 40 - sqrt(2) * w --- (1)

Area of trapezium A = 1/2 * L * height
Height of the trapezium = w * sin(theta) = w * sin(45) = w / sqrt(2)
A = 1/2 * L * w / sqrt(2)
Substitute L from (1) and you will get A as a function of w.
To maximize A, find dA/dw,, equate to 0, solve for w.

Answer 2

Just noticed a mistake in equation (1). It should be 2w and not 2wcos(45).
Width of sheet metal = 2 * w + L = 40
L = 40 - 2w --- (1)

Answer 3

isnt the area of a trapezium 1/2(a+b)*h?

Answer 4

@ranga ??

Answer 5

Oh man you are right. Too tired and hence many human errors.

Answer 6

The length of the top side is: (2wcos(45) + L) = 2 * w / sqrt(2) + L
Area of trapezium A = 1/2 * (L + 2*w/sqrt(2) + L) * w / sqrt(2)
A = 1/2 * (2L + 2*w/sqrt(2)) * w/sqrt(2) = (L + w/sqrt(2)) * w /sqrt(2)
A = Lw/sqrt(2) + w^2/2
Put L = 40 - 2w in A.
A = w/sqrt(2) * (40 - 2w) + w^2/2
A = 40w/sqrt(2) - sqrt(2)w^2 + w^2/2
A = 40w/sqrt(2) + w^2(1/2 - sqrt(2))

dA/dw = 40/sqrt(2) + 2w(1/2-sqrt(2)) = 0

find w.