Question

A metal channel is formed by turning up the sides of width x of a rectangular sheet of metal through an angle theta. If the sheet is 200mm wide, determine the values of x and theta for which the cross-section of the channel will be a maximum?

Cannot picture it. What initial formulae will be obtained and how are they obtained?

Answer 1

Cross Sectional Area `= 5000 \ mm^2`

Explanation:

Let us setup the following variables:

# { (x, "width of sheet","(mm)"), (theta, "Angle between folded sides", "(radians)"; 0 lt theta lt pi), (A, "Cross Sectional Area", "(mm"^2")") :} #

The width of the sheet is `x`, so when folded the cross-sectional area is a triangle with sides `1/2x` separated by an angle `theta`, which we measure in radians in preparation for calculus operations. The area of the triangle is therefore:

`A = (1/2)(1/2x)(1/2x)sin theta`
`\ \ \ = 1/8x^2sin theta`

We are given that `x=200` giving us:

`A = (1/2)(1/2x)(1/2x)sin theta`
`\ \ \ = 20000/8sin theta`
`\ \ \ = 5000sin theta`

Differentiating wrt `theta`, we have:

`(dA)/(d theta) = 5000cos theta`

At a maximum/minimum, the derivative vanishes, giving us:

`5000cos theta = 0 => theta = pi/2`

With this value of `theta` we find that:

`A = 5000sin (pi/2) = 5000`