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...

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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*