a. A textbook company discovered that the profit made from selling its books is given by y = \(\frac{x^2}{8}\) + 5x, where x is the number of textbooks sold (in thousands) and y is the corresponding profit (in Ghana Cedis). If the company made a profit of GH₵ 20,000.00
i. form a quadratic equation in x;
ii. (using the quadratic formula, find, correct to the nearest whole number, the number of textbooks sold to make the profit.
b. The angle of elevation of the top T of a tree from a point P on the same ground level as the foot Q of a tree is 28\(^o\). A bird perched at a point R, halfway up the tree.
i. Represent the information in a diagram.
ii. Calculate, correct to the nearest degree, the angle of elevation of R from P.
(a)(i) With profit \(y = 20000\):
\[\frac{x^2}{8} + 5x = 20000 \;\;(\times 8) \Rightarrow x^2 + 40x - 160000 = 0\]
(a)(ii) Quadratic formula:
\[x = \frac{-40 \pm \sqrt{40^2 + 4(160000)}}{2} = \frac{-40 \pm \sqrt{641600}}{2} = \frac{-40 \pm 801.0}{2}\]
Taking the positive root: \(x = \dfrac{761.0}{2} \approx 380\) (in thousands), i.e. about \(380{,}000\) textbooks.
(b)(i) Diagram: vertical tree \(TQ\) with P on the ground so that \(\angle TPQ = 28^{\circ}\); R is the midpoint of \(TQ\).
(b)(ii) Let \(PQ = d\). Then \(TQ = d\tan 28^{\circ}\) and \(RQ = \tfrac{1}{2}TQ = \tfrac{1}{2}d\tan 28^{\circ}\).
\[\tan(\angle RPQ) = \frac{RQ}{PQ} = \tfrac{1}{2}\tan 28^{\circ} = \tfrac{1}{2}(0.5317) = 0.2659\]
\[\angle RPQ = \tan^{-1}(0.2659) \approx 15^{\circ}\]
(a)(i) With profit \(y = 20000\):
\[\frac{x^2}{8} + 5x = 20000 \;\;(\times 8) \Rightarrow x^2 + 40x - 160000 = 0\]
(a)(ii) Quadratic formula:
\[x = \frac{-40 \pm \sqrt{40^2 + 4(160000)}}{2} = \frac{-40 \pm \sqrt{641600}}{2} = \frac{-40 \pm 801.0}{2}\]
Taking the positive root: \(x = \dfrac{761.0}{2} \approx 380\) (in thousands), i.e. about \(380{,}000\) textbooks.
(b)(i) Diagram: vertical tree \(TQ\) with P on the ground so that \(\angle TPQ = 28^{\circ}\); R is the midpoint of \(TQ\).
(b)(ii) Let \(PQ = d\). Then \(TQ = d\tan 28^{\circ}\) and \(RQ = \tfrac{1}{2}TQ = \tfrac{1}{2}d\tan 28^{\circ}\).
\[\tan(\angle RPQ) = \frac{RQ}{PQ} = \tfrac{1}{2}\tan 28^{\circ} = \tfrac{1}{2}(0.5317) = 0.2659\]
\[\angle RPQ = \tan^{-1}(0.2659) \approx 15^{\circ}\]
