Nov 2018 Q2
a) What is a Stationary Point of a differentiable function of one variable? (2 marks)
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A stationary point of a differentiable function of one variable is a point (of the domain) of the function where the derivative is zero. Equivalently, the slope of the graph at that point is zero.
b) Renes Trading Company sells q kente strips per month at p Ghana Cedis per Kente strip. The demand function for kente strips is given by p=300−0.02q. The kente strips cost GH¢30 per strip to manufacture. There are fixed costs of GH¢9,000 per month.
Required:
i) Determine the price per kente strip that will maximize revenue. (4 marks)
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The demand function is given by p=300−0.02q
Thus the Total Revenue, TR(q) = (300−0.02q)q
The critical points of revenue is obtained by taking the derivative of TR(q)
Thus revenue is maximum at q=7500
Price that maximizes revenue is p=300=0.02(7500) = 150 GH₵
ii) Determine the quantity where profit is maximized. (4 marks)
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iii) Calculate the maximum profit. (2 marks)
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c) Due to changes in market conditions, the company finds the demand q, in thousands, for their kente strips to be at a price of GH¢p (p Ghana Cedis).
Required:
i) Determine the elasticity of demand when the price is GH¢5 and when the price is GH¢15 per kente strip. (6 marks)
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ii) Comment on your results in (i). (2 marks)
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It is inelastic at price 5 and elastic at 15.