# Thread: More Optimization Problem . . .

1. ## More Optimization Problem . . .

A painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How far from the wall should the observer stand to get the best view? (In other words, where should the observer stand to maximize the angle theta subtended at his eye by the painting?)

I'm confident that I've worked the first part of the problem correctly, so I will just post my work for the part that I am having trouble with.

theta = a and beta = b

a = (a + b) - b
a = tan^-1(h + d/x) - tan^-1(d/x)
Da/dx = -(h + d)/(x^2 + h^2 + 2dh + d^2) +( )d/(x^2 + d^2)
=> -(h + d)(x^2 +d^2) + (d)(x^2 + h^2 + 2dh + d^2)/(x^2 + h^2 + 2dh + d^2)(x^2 + d^2)

Thanks

2. Hello, zachb!

I agree with your set-up . . . then I went off on a different path.

A painting in an art gallery has height h and is hung
so that its lower edge is a distance d above the eye of an observer.
(In other words, where should the observer stand to maximize
the angle theta subtended at his eye by the painting?)
Code:
                                    * B
*   |
*       | h
*           |
*               * C
*           *       |
*       *               | d
* θ   *                     |
A * - - - - - - - - - - - - - - * D
x

We have: .θ .= .arctan[(h+d)/x] - arctan(d/x)

. . which equals: .θ .= .arccot[x/(h+d)] - arccot(x/d)

. . . . . . . . . . . - 1/(h+d) . . . . . 1/d
Then: .θ' .= .----------------- + ------------ .= .0
. . . . . . . . . 1 + [x/(h+d)]² . .1 + (x/d)²

Multiply the first fraction by (h+d)²/(h+d)², the second by d²/d²

. . . . . . . . . . . . . . -(h+d) . . . . . . d
. . and we have: .--------------- + ---------- .= .0
. . . . . . . . . . . . .(h+d)² + x² . . d² + x²

Clear denominators: .-(h+d)(x² + d²) + d[x² + (h+d)²] .= .0

Expand (partially): .-(h+d)x² - d²(h+d) + dx² + d(h+d)² .= .0

. . and we have: .dx² - (h+d)x² .= .d²(h+d) - d(h+d)²

Factor: .[d - (h+d)]·x² .= .d(h+d)·[d - (h+d)]

. . and we have: .-hx² .= .-hd(h+d) . . .= .d(h+d)

. . . . . . . . . . . . . _______
Therefore: .x .= .√d(h + d)

Someone check my work . . . please!

3. We have: .θ .= .arctan[(h+d)/x] - arctan(d/x)

. . which equals: .θ .= .arccot[x/(h+d)] - arccot(x/d)
This step seems a litte strange to me. Why did you go from arctan to arccot?

,

,

### a painting in an art gall

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