A gutter is to be made using a 5m long rectangular piece of metal that has to be bent to form the open topper gutter. the cross section of the gutter is an isosceles trapezoid with sides making angles 120 degrees with the base?

If the top width of the gutter is 60 cm, find the dimensions so that the cross sectional area will be a maximum.

1 Answer
Barry H.
Jun 24, 2018

![enter image source here]The cross section of the sheet required would have a base length of #30cm# and sides of the same length.(https:

Explanation:

Let the base of the isosceles trapezoid have length #x# and height of #h#. From the sketch it can be seen that #a=[[60-x]]/2#...........#[1]#

Considering one of the isosceles triangles formed at the base of the large isosceles triangle it is seen that #tan 30 =a/h#.

#tan 30=sqrt3/3#, therefore, from .....#[1]# , #h=3[[60-x]]/[2sqrt3#..........#[2]#

Since the base angle of the trapezoid is constant, the area of the rectangle contained within the large isosceles triangle must be maximised.

Area of rectangle= #xh# = #x[3[60-x]]/[2sqrt3# = #[180x/sqrt3-3x^2/sqrt3]#.........#[3]#

For max/min, differentiating Area, say #A# wrtx.
#[dA]/[dx] =180/[2sqrt3]-6x/[2sqrt3]# = 0 [ for max/min]. From this it is shown that when #x=30# the area will be maximised.

Checking the second derivative which will be negative for a max turning point, #d^2A/dx^2# =#-6/2sqrt3#, which is negative , whatever the value of #x#.

From ........#[2]# when #x=30#, #h= 3[60-30]/[2sqrt3# = #45sqrt3/3#. And from .....#[1]# when #x = 30#, #a= 15#

By Pythagoras, the side length #L# of the small isosceles triangle, which is the side length of the gutter,

#15^2+[45sqrt3/3]^2=L^2#, #L# = #sqrt900 # =#30#.

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