A hoarding of height 5 m is mounted on a vertical wall. The bottom of the hoarding is at a height of 4 m above the horizontal eye level of an observer. The distance from the wall at which the observer must stand so that the hoarding subtends the maximum possible angle at their eye is:

Step-by-step solution

Let the distance of the observer from the wall be x . Let the angle of elevation to the top of the hoarding be and to the bottom of the hoarding be . The height of the top of the hoarding from eye level is 4 + 5 = 9 m, and the height of the bottom is 4 m. Therefore, = 9 x and = 4 x . The angle subtended by the hoarding at the observer's eye is = - . = ( - ) = - 1 + = 9 x - 4 x 1 + ( 9 x ) ( 4 x ) = 5x x^2 + 36 To maximize , we must maximize (since is acute). = 5 x + 36 x To maximize this expression, the denominator