# A gutter with an open top and a rectangular cross section is to be formed from a sheet of metal 40 cm wide by bending up equal strips(walls)..........along the edges. In order to carry the maximum...

A gutter with an open top and a rectangular cross section is to be formed from a sheet of metal 40 cm wide by bending up equal strips(walls).....

.....along the edges. In order to carry the maximum amount of water, the area formed by the walls and the floor of the gutter must be at its maximum possible value. What is the height of the walls and the width of the floor?

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Width of the metal sheet = 40cm

When you form a open top rectangular gutter using this metal sheet you will have a width of floor and two equal vertical wall heights.

Let;

width of floor `= y`

Vertical wall height `= x`

So we can write;

`y+x+x = 40`

`y+2x = 40`

`y = 40-2x`

If Cross sectional area of the gutter is A;

`A = x*y`

In order to carry maximum volume we should have the maximum cross section in the gutter. At maximum cross section,the derivative of A will become 0.

`A = x*y`

`A = x(40-2x)`

`A = 40x-2x^2`

`(dA)/dx = 40-4x`

When `(dA)/dx = 0` ;

`40-4x = 0`

` x = 10`

If A is a maximum then `(d^2A)/dx^2 < 0`

`(d^2A)/dx^2= -4 < 0 `

So we have maximum for the gutter area.

*Maximum area of the gutter is given when;*

*Width of floor = 20*

*Wall height = 10*

*Maximum area of gutter (A) = 10*20 = 200cm^2*