Answer to Question #144530 in Calculus for Nawar Ariza Choudhuri

Question #144530

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


1
Expert's answer
2020-11-17T06:17:57-0500

Let man is at a distance of xx from the wall.

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

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

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


Then,

tan(θ+α)=h+dxtan(\theta + \alpha) = \frac{h+d}{x}

Differentiating both sides,

sec2(θ+α)(θ(x)+α(x))=h+dx2sec^2(\theta + \alpha)(\theta'(x) + \alpha'(x))=-\frac{h+d}{x^2}


θ(x)+α(x)=(h+d)cos2(θ+α)x2\theta'(x) + \alpha'(x)=-\frac{(h+d)cos^2(\theta + \alpha)}{x^2}


Putting value of cos(θ+α)=xx2+(d+h)2cos(\theta+\alpha) = \frac{x}{\sqrt{x^2+(d+h)^2}}


then,

θ(x)+α(x)=(h+d)x2x2(x2+(d+h)2)=h+dx2+(d+h)2\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,

tanα=dx    α=tan1dxtan \alpha = \frac{d}{x} \implies \alpha =tan^{-1} \frac{d}{x}


Differentiating it, we get,

α(x)=x2d2+x2dx2=dd2+x2\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,

θ(x)=h+dx2+(d+h)2α(x)=dd2+x2h+dx2+(d+h)2\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

θ(x)=h+dx2+(d+h)2α(x)=dd2+x2h+dx2+(d+h)2=0\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+dx2=(h+d)(d2+x2)d(h+d)^2+dx^2=(h+d)(d^2+x^2)

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


Putting value of d and h in above equation,

x=2mx = 2 m





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