Six problems in maxima and minima

Equate to zero, and, x — a that is to say, the line a must be divided into two equal parts, and the greatest possible rectangle is a square. 

It is the rule, when seeking maxima and minima, to simplify the process by omitting the constant factors, since, whatever makes the variable hx - x2 a maximum will also make b(hx - x2)/h a maximum.1 Now differentiate the expression obtained above for the area of the rectangle neglecting bjh, and equate the result to zero, in this way we obtain 

That is to say, the height of the rectangle must be half the altitude of the triangle. 

To cut a sector from a circular sheet of metal so that the remainder can be formed into a conical-shaped vessel of maximum capacity. Let ACB (Fig. 75) be a circular plate of radius, r, it is required to cut out a portion A OB such that the conical vessel formed by joining OA and OB together may hold the greatest possible amount of fluid. Let x denote the angle remaining after the sector AOB has been removed. We must first find a relation between x and the volume, v, of the cone.2 

1 This is easily proved, for let y = cf (x), where c has any arbitrary constant value. For a maximum or minimum value dy/dx = cf x) = 0, and this can only occur where 

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