Question

Question #b1a21

Answer 1

That is correct. See explanation.

Explanation:

First of all, we need to figure out the expression that relates `h`, `c`, `theta` and `x`.
Let's call the observer's point O, the point on the wall along the horizontal line-of-sight (at the bottom of `c`) P, the point at the bottom of the art figure A, and the top of the figure B.
So we have three triangles OPA and OAB, together forming OPB.

Let's also call the angle AOP `alpha` and the grand angle BOP `beta` so that `theta = beta - alpha`.

We know that:
`tan alpha = c/x` and `tan beta = (h+c)/x`
So
`alpha = arctan (c/x)` and `beta = arctan ((h+c)/(x))`

Rewriting everything, we have:
`theta = arctan ((h+c)/(x)) - arctan (c/x)`

We take the derivative of `theta` to `x` and get:
`(d theta)/(d x) = (-(h+c)/(x^2))/(1+((h+c)/(x))^2) - (-c/(x^2))/(1+ (c/x)^2)`
because the derivative of `d/(d k) arctan k = (1)/(1+k^2)`.

Rearranging, we get:
`(d theta)/(d x) = (-(h+c)/(x^2))/((x^2+(h+c)^2 )/(x^2)) - (-c/(x^2))/((x^2+ c^2)/x^2)`
i.e.:
`(d theta)/(d x) = (-(h+c))/(x^2+(h+c)^2 ) - (-c)/((x^2+ c^2)`
i.e.:
`(d theta)/(d x) = (-(h+c) (x^2+ c^2))/((x^2+(h+c)^2 )(x^2+ c^2)) - (-c (x^2+(h+c)^2 ))/((x^2+(h+c)^2 )(x^2+ c^2))`

Set the derivative to zero and solve for x.
As you can see, the denominator is always positive, so, we only need to solve for the numerator to be zero.
In the numerator, we can factor out the `x^2`.
So, we are left with:
`0 = -hx^2 -cx^2 -hc^2-c^3+ cx^2 +ch^2 + 2c^2h +c^3`
i.e.
`x^2 = ch+2c = c(h+c)`
i.e.
`x = sqrt(c(h+c))`
(because we omit the answer that is negative since we are talking about a distance, which is always positive).
Q.E.D.

Answer 2

See below.

Explanation:

We have

`tan(theta+alpha) = (h+c)/x = (tan theta + tan alpha)/(1-tan theta tan alpha)` and solving for `tan theta`

`tan theta = (c + h - x tanalpha)/(x + c tanalpha + h tanalpha)`

but `tan alpha = c/x` and after substituting

`tan theta = (x h)/(x^2 + c (c + h))`

then `tan theta(x)` is maximum for

`d/(dx)tan theta(x) = (h (c^2 - x^2 + c h))/(c^2 + x^2 + c h)^2=0`

which occurs for `x = sqrt[c(c + h)]`

and then

`theta_(max) = arctan(sqrt[c(c + h)])`

NOTE

`tan theta` is monotonically increasing for `0 < theta < pi/2`