Science:Math Exam Resources/Courses/MATH102/December 2014/Question C 03/Solution 1

As indicated in the hint, we define $\theta$ as the angle to the top of the screen* and $\phi$ as the angle to the bottom of the screen. Then $\tan(\theta )={\frac {8}{x}},\quad \tan(\phi )={\frac {2}{x}}.$ Maximizing the visual angle occupied by the screen is equivalent to maximizing the angle $\theta -\phi$. As $\theta$ and $\phi$ are both functions of $x$, we just need to calculate the derivative $\theta '-\phi '$ and find the critical $x$. First with implicit differentiation we calculate

Solution!

${\begin{aligned}(\tan(\theta ))'&=({\frac {8}{x}})'\\\theta '\sec ^{2}(\theta )&=-{\frac {8}{x^{2}}}\\\theta '&=-{\frac {8}{x^{2}}}\cos ^{2}(\theta ).\end{aligned}}$

Note: It is a common mistake to interpret the angle $\phi$ as the angle that is below the angle $\theta$ and not the angle that is within the angle $\theta$.

With the diagram on the right

we obtain $\cos(\theta )={\frac {x}{\sqrt {8^{2}+x^{2}}}}.$ So $\theta '=-{\frac {8}{x^{2}}}{\frac {x^{2}}{8^{2}+x^{2}}}=-{\frac {8}{8^{2}+x^{2}}}.$ Similarly, we calculate $\phi '=-{\frac {2}{2^{2}+x^{2}}}.$ Hence,

$\theta '-\phi '=-{\frac {8}{8^{2}+x^{2}}}+{\frac {2}{2^{2}+x^{2}}}={\frac {-8(4+x^{2})+2(64+x^{2})}{(64+x^{2})(4+x^{2})}}={\frac {-6x^{2}+96}{(64+x^{2})(4+x^{2})}}.$

By setting $\theta '-\phi '=0$ we solve $x^{2}=96/6=16$, $x=4$ as $x$ represents the distance and must be positive. To verify it is a maximum, we use the first derivative test

${\begin{aligned}\theta '-\phi '{\Big |}_{x=3}&={\frac {-6\cdot 9+96}{(64+3^{2})(4+3^{2})}}={\frac {96-54}{(64+3^{2})(4+3^{2})}}>0,\\\theta '-\phi '{\Big |}_{x=5}&={\frac {-6\cdot 25+96}{(64+5^{2})(4+5^{2})}}={\frac {96-150}{(64+5^{2})(4+5^{2})}}<0.\\\end{aligned}}$

So the angle is maximized at $x=4$.