# Question b1a21

Feb 15, 2018

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:
$\tan \alpha = \frac{c}{x}$ and $\tan \beta = \frac{h + c}{x}$
So
$\alpha = \arctan \left(\frac{c}{x}\right)$ and $\beta = \arctan \left(\frac{h + c}{x}\right)$

Rewriting everything, we have:
$\theta = \arctan \left(\frac{h + c}{x}\right) - \arctan \left(\frac{c}{x}\right)$

We take the derivative of $\theta$ to $x$ and get:
$\frac{d \theta}{d x} = \frac{- \frac{h + c}{{x}^{2}}}{1 + {\left(\frac{h + c}{x}\right)}^{2}} - \frac{- \frac{c}{{x}^{2}}}{1 + {\left(\frac{c}{x}\right)}^{2}}$
because the derivative of $\frac{d}{d k} \arctan k = \frac{1}{1 + {k}^{2}}$.

Rearranging, we get:
$\frac{d \theta}{d x} = \frac{- \frac{h + c}{{x}^{2}}}{\frac{{x}^{2} + {\left(h + c\right)}^{2}}{{x}^{2}}} - \frac{- \frac{c}{{x}^{2}}}{\frac{{x}^{2} + {c}^{2}}{x} ^ 2}$
i.e.:
(d theta)/(d x) = (-(h+c))/(x^2+(h+c)^2 ) - (-c)/((x^2+ c^2)#
i.e.:
$\frac{d \theta}{d x} = \frac{- \left(h + c\right) \left({x}^{2} + {c}^{2}\right)}{\left({x}^{2} + {\left(h + c\right)}^{2}\right) \left({x}^{2} + {c}^{2}\right)} - \frac{- c \left({x}^{2} + {\left(h + c\right)}^{2}\right)}{\left({x}^{2} + {\left(h + c\right)}^{2}\right) \left({x}^{2} + {c}^{2}\right)}$

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:
$0 = - h {x}^{2} - c {x}^{2} - h {c}^{2} - {c}^{3} + c {x}^{2} + c {h}^{2} + 2 {c}^{2} h + {c}^{3}$
i.e.
${x}^{2} = c h + 2 c = c \left(h + c\right)$
i.e.
$x = \sqrt{c \left(h + c\right)}$
(because we omit the answer that is negative since we are talking about a distance, which is always positive).
Q.E.D.

Feb 15, 2018

See below.

#### Explanation:

We have

$\tan \left(\theta + \alpha\right) = \frac{h + c}{x} = \frac{\tan \theta + \tan \alpha}{1 - \tan \theta \tan \alpha}$ and solving for $\tan \theta$

$\tan \theta = \frac{c + h - x \tan \alpha}{x + c \tan \alpha + h \tan \alpha}$

but $\tan \alpha = \frac{c}{x}$ and after substituting

$\tan \theta = \frac{x h}{{x}^{2} + c \left(c + h\right)}$

then $\tan \theta \left(x\right)$ is maximum for

$\frac{d}{\mathrm{dx}} \tan \theta \left(x\right) = \frac{h \left({c}^{2} - {x}^{2} + c h\right)}{{c}^{2} + {x}^{2} + c h} ^ 2 = 0$

which occurs for $x = \sqrt{c \left(c + h\right)}$

and then

${\theta}_{\max} = \arctan \left(\sqrt{c \left(c + h\right)}\right)$

NOTE

$\tan \theta$ is monotonically increasing for $0 < \theta < \frac{\pi}{2}$