Solution — 2023 Question 5 Differentiation & Applications

(a)

Considering the perimeter,

\begin{matrix} \frac{1}{2} (2 π x) + 2 y + 2 x & = 4 \\ π x + 2 y + 2 x & = 4 \\ 2 y & = 4 - π x - 2 x \\ y & = 2 - \frac{1}{2} π x - x \end{matrix}

\begin{matrix} A & = 2 x y + \frac{1}{2} (π x^{2}) \\ = 2 x (2 - \frac{1}{2} π x - x) + \frac{1}{2} (π x^{2}) \\ = 2 x (2 - \frac{1}{2} π x - x) + \frac{1}{2} π x^{2} \\ = 4 x - x^{2} π - 2 x^{2} + \frac{1}{2} π x^{2} \\ = 4 x - \frac{1}{2} x^{2} π - 2 x^{2} \\ = 4 x - \frac{1}{2} x^{2} (π + 4) ∎ \end{matrix}

(b)

\begin{matrix} A & = 4 x - \frac{1}{2} x^{2} (π + 4) \\ \frac{d A}{d x} & = 4 - \frac{1}{2} (π + 4) (2 x) \\ = 4 - (π + 4) x \end{matrix}

At maximum value of $A,$ $\frac{d A}{d x} = 0$

\begin{matrix} 4 - (π + 4) x & = 0 \\ (π + 4) x & = 4 \\ x & = \frac{4}{π + 4} \end{matrix}

\begin{matrix} A & = 4 (\frac{4}{π + 4}) - \frac{1}{2} {(\frac{4}{π + 4})}^{2} (π + 4) \\ = \frac{16}{π + 4} - \frac{8}{π + 4} \\ = \frac{8}{π + 4} ∎ \end{matrix}

\begin{matrix} \frac{d A}{d x} & = 4 - (π + 4) x \\ \frac{d^{2} A}{d x^{2}} & = - (π + 4) \end{matrix}

Since $\frac{d^{2} A}{d x^{2}} < 0,$ $A = \frac{8}{π + 4}$ is a maximum $∎$

(c)

Considering the perimeter,

\begin{matrix} 5 (2 z) & = 4 \\ 10 z & = 4 \\ z & = \frac{2}{5} \end{matrix}

Let $h$ be the height of the triangle. By Pythagoras’ Theorem,

\begin{matrix} h & = \sqrt{{(2 z)}^{2} - z^{2}} \\ = z \sqrt{3} \\ = \frac{2}{5} \sqrt{3} \\ = 0.692 82 \end{matrix}

\begin{matrix} Area of glass \\ = {(2 z)}^{2} + \frac{1}{2} (2 z) h \\ = {(2 (\frac{2}{5}))}^{2} + \frac{1}{2} (2 (\frac{2}{5})) (0.692 82) \\ = 0.917 13 \\ = 0.917 ∎ m^{2} (to 3 s.f.) \end{matrix}

(d)

Considering Type $X,$

\begin{matrix} C + \frac{8}{π + 4} D & = 103 \\ C + 1.1202 D & = 103 & (1) \end{matrix}

\begin{matrix} 0.8 C + 0.917 13 D & = 83 & (2) \end{matrix}

Solving (1) and (2) with a GC,

$C = 70.9 ∎ and D = 28.6 ∎$