A gutter is to be made from a piece of sheet metal that is aa inches wide by bending up pieces of equal width at a right angle. Determine the width (xx) of the piece the needs to be bent up to maximize the cross section of the gutter.

The answer to the question is

The width (x) (the middle portion) of the piece that needs to be bent up to maximize the cross sectional area of the gutter is x = a/2 inches.

Step-by-step explanation:

To solve the question we note that this is a word problem

Let the width (base) of the gutter be = x

The height of the gutter = b

Therefore the perimeter = a = x + 2·b

or x = a - 2·b

The area A = bx = (a - 2·b)×b = ab -2·b²

to find the local maximum, we differentiate the equation for the area and we equate it to zero to get the extremums

Therefore A' = a - 4·b = 0 or a = 4·b and b = a/4

To verify if this value is a local maximum or minimum, we differentiate again to get A'' = -4 which means the function is concave down with no minimum value this gives b = a/4 as a local maximum

Since a/4 is a local maximum we have height of gutter, b = a/4 and the base x =a - 2·b = a - a/2 = a/2 and therefore the area is given by

a/2*a/4 = a²/8

Note If we however we note that for a special case of a gutter the top can be left open so we have three equal segments with height to be x = a/3

we have area = a/3 × a/3 = a²/9 which is lesser than a²/8

Therefore the width (x) of the piece the needs to be bent up to maximize the cross section of the gutter is x = a/2 inches

1/4 and 8/3 because:

the reciprocal if 4 is 1/4 because 4 is used as 4/1 and the opposite of that is 1/4. the opposite of 3/8 is 8/3 so that is the answer.