Question

1. A company estimates that if r thousand cedis is spent on the marketing of a certain product, Q(1) 7 thousand units of the product will be sold, where Q() 27 +22 (a) Dctcrmine the domain of the function defined. 12 Marks] (b) Determine how much of the product will be sad if the company does not spend any money on marketing 12 Marks/ (c) Determine the amount of money the company must spend on marketing to achieve the minimum and maximum quantities. Leave your answers to 2 decimal places. 110 Marks) (d) Determine the maximum and minimum quantities that can be sold. Leave your answer to the nearest whole number. 15 Marks/ (e) Determine the marketing expendure ranges on which quantity sold is increasirg and decreasing and interpret your solution. 19 Marks/ (f) Determine what happens as the company invests an infintely large amount of money on market- ing. 16 Marks/ (g) Use the information given from (a) to (f) to sketch the curve of Q(r). 16 Marks) 2. (a) The demand and supply functions of a particular good in a competitive market are defined respectively as y = 2007- and p = 30+2. If pand denote the price and quantity respectively, determine the equilibrium price and quantity. 115 Marks/ (b) Let the price and quantity functions of a product be given by p = 2r + k and q = 1+2 respectively, where x is a variable ard k is a constant. Find the value(s) of k for which the total revenue function is clways positive. 115 Marks/ Examiners: G. A. Botchway, L. F. Kyei Page 1 of 2 Section B 3. (a) Suppose the Gross Domestic Product (GDP) projection for Ghara is given by the function G(t)= 250 – 100e-051 million cedis, where t is the number of years after the year 2009. i. Calculate the GDP in the year 2019. 14 Marks/ i. What is the rate of change of GDP with respect to time in the year 2019? 18 Marks) iii. At what percentage is the GDP changing with respect to time in the year 2019? Leave your answer to two decimal place. (5 Marks) (b) Solve the equation 223 – 6.24 + 8 =D 13 Marks/

Answer 1

1. Q(1) 7.0 27 +22

(a) Domain is r ER, function is defined for all real values of r

(b) Q(0) = 0, no product will be sold

(c), (d) For maxima and minima, equate derivative to zero

Q'(x) = d 7.0 dr 27 +22 7(27 – 1?) (27+.x2)

$Q'(x)=0~\Rightarrow 27-x^{2}=0~\Rightarrow x=\sqrt{27}\approx 5.196$

When I = 5.196, Q(5.196) 0.674

$\textup{Hence, maximum } 674\textup{ units are sold when }5196 \textup{ cedis are spent on marketing}$

$\textup{The minimum product sold is 0, when 0 cedis are spent }$

$\textup{(e) }Q'(x)=\frac{7(27-x^{2})}{(27+x^{2})^{2}}>0\textup{ when }x<\sqrt{27}\approx 5.196$

Hence, quantity sold is increasing when upto 5196 cedis are spent and

$\textup{decreasing when more than 5196 cedis are spent. It implies that }$

$\textup{product sales increases only for marketing expense upto 5196 cedis }$

$\textup{(f) }\lim_{x\rightarrow \infty }Q(x)=\lim_{x\rightarrow \infty }\frac{7x}{27+x^{2}}=0$

$\textup{Hence, no product will be sold if an infinitely large amount is spent on marketing}$

$\textup{(g)}$

1 (5.196, 0.6736) 0:54 -5 10 15 20 25 30 35 40 0 5

$\textbf{2. }\textup{ (a) Under equilibrium, demand and supply are equal. Hence,}$

$\frac{200}{q}=30+2q$

92 + 159 – 100 = 0

$(q+20)(q-5)=0$

$\textup{Since quantity is positive, hence equilibrium quantity }q=5$

$\textup{and equilibrium price}=\frac{200}{5}=40$

$\textup{ (b) Revenue, }R(x)=\textup{Price}\times \textup{Quantity}=(2x+k)(x+2)$

$R(x)=2x^{2}+(k+4)x+2k\textup{ will be always positive, if its discriminant is negative}$

$\textup{Hence, }D=(k+4)^{2}-4\times 2\times 2k<0$

$k^{2}-8k+16<0$

$(k-4)^{2}<0$

$\textup{A perfect square can never be negative, hence no value of }k\textup{ is feasible}$

$\textbf{3. }G(t)=250-100e^{-0.5t}$

$\textup{i. In 2019, }t=0.\textup{ Hence, }G(10)=250-100e^{-5}\approx 249.3262\textup{ million cedis}$

$\textup{ii. Rate of change, }G'(t)=50e^{-0.5t}$

$\textup{Hence, }G'(10)=50e^{-5}\approx 0.3369\textup{ million cedis/year}$

$\textup{iii. Percentage change}=\frac{G'(t)}{G(t)}\times 100\approx 0.14\%$

$\textup{(b) }\textup{Let }y=2^{x},\textup{ then equation can be w=rewritten as}$

$y^{2}-6y+8=0$

$(y-2)(y-4)=0$

$y=2^{x}=2,4$

$\textup{Hence, solutions are }x=1,2$