Answer to Question #144530 in Calculus for Nawar Ariza Choudhuri

Question

A painting in an art gallery has height 3 m and is hung so that its lower edge is about 1 m above the eye of an observer. How far from the wall should the observer stand to get the best view? (i.e. the observer wants to maximize the angel θ subtended at the eye by the painting

Answer 1

Let man is at a distance of $x$ from the wall.

Let angle made made at lower and upper end be $\alpha$ and $\theta$ respectively.

for sake of simplicity, Let height of the painting is $h=3m$

and height of the lower end from the ground is, $d=1m$

Then,

$tan(\theta + \alpha) = \frac{h+d}{x}$

Differentiating both sides,

$sec^2(\theta + \alpha)(\theta'(x) + \alpha'(x))=-\frac{h+d}{x^2}$

$\theta'(x) + \alpha'(x)=-\frac{(h+d)cos^2(\theta + \alpha)}{x^2}$

Putting value of $cos(\theta+\alpha) = \frac{x}{\sqrt{x^2+(d+h)^2}}$

then,

$\theta'(x) + \alpha'(x)=-\frac{(h+d)x^2}{x^2(x^2+(d+h)^2)} = -\frac{h+d}{x^2+(d+h)^2}$ (1)

Now,

$tan \alpha = \frac{d}{x} \implies \alpha =tan^{-1} \frac{d}{x}$

Differentiating it, we get,

$\alpha'(x) = \frac{x^2}{d^2+x^2}\frac{-d}{x^2} = -\frac{d}{d^2+x^2}$ (2)

From (1) and (2),

we get,

$\theta'(x) = -\frac{h+d}{x^2+(d+h)^2}- \alpha'(x) = \frac{d}{d^2+x^2}-\frac{h+d}{x^2+(d+h)^2}$

To maximize the area, put

$\theta'(x) = -\frac{h+d}{x^2+(d+h)^2}- \alpha'(x) = \frac{d}{d^2+x^2}-\frac{h+d}{x^2+(d+h)^2} = 0$

Solving it, we get,

$d(h+d)^2+dx^2=(h+d)(d^2+x^2)$

$x = \sqrt{d(d+h)}$

Putting value of d and h in above equation,

$x = 2 m$