Solution — 2017 Question 3 Differentiation & Applications

(a)

Let $L$ cm be the length of the rectangle.

Perimeter of sign:

\begin{matrix} 2 L + 2 π x & = 10 \\ 2 L & = 10 - 2 π x \\ L & = 5 - π x \end{matrix}

Area of sign:

\begin{matrix} A & = 2 x L + π x^{2} \\ = 2 x (5 - π x) + π x^{2} \\ = 10 x - 2 x^{2} π + π x^{2} \\ = 10 x - x^{2} π \\ = x (10 - x π) ∎ \end{matrix}

(b)

\begin{matrix} A & = x (10 - π x) \\ = 10 x - x^{2} π \\ \frac{d A}{d x} & = 10 - π (2 x) \\ = 10 - 2 π x \end{matrix}

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

\begin{matrix} 10 - 2 π x & = 0 \\ 2 π x & = 10 \\ x & = \frac{5}{π} ∎ \end{matrix}

When $x = \frac{5}{π},$

\begin{matrix} A & = x (10 - π x) \\ = \frac{5}{π} (10 - π (\frac{5}{π})) \\ = \frac{25}{π} ∎ {cm}^{2} \end{matrix}

\begin{matrix} \frac{d A}{d x} & = 10 - 2 π x \\ \frac{d^{2} A}{d x^{2}} & = - 2 π \end{matrix}

Since $\frac{d^{2} A}{d x^{2}} < 0,$ the area is a maximum. $∎$