The fencing used for side AB costs \(\$ 11\) per metre. The fencing for the other three sides costs \(\$ 3\) per metre. The farmer creates an enclosure so that the cost is a minimum. Find this minimum cost.

METHOD 1

correct expression for second side, using area = 525     (A1)

e.g. let \({\rm{AB}} = x\) ,  \({\rm{AD}} = \frac{{525}}{x}\)

attempt to set up cost function using $3 for three sides and $11 for one side     (M1)

e.g. \(3({\rm{AD}} + {\rm{BC}} + {\rm{CD}}) + 11{\rm{AB}}\)

correct expression for cost     A2

e.g. \(\frac{{525}}{x} \times 3 + \frac{{525}}{x} \times 3 + 11x + 3x\) , \(\frac{{525}}{{{\rm{AB}}}} \times 3 + \frac{{525}}{{{\rm{AB}}}} \times 3 + 11{\rm{AB}} + 3{\rm{AB}}\) , \(\frac{{3150}}{x} + 14x\)

EITHER

sketch of cost function     (M1)

identifying minimum point     (A1)

e.g. marking point on graph, \(x = 15\)

minimum cost is 420 (dollars)     A1     N4

OR

correct derivative (may be seen in equation below)     (A1)

e.g. \(C'(x) = \frac{{ - 1575}}{{{x^2}}} + \frac{{ - 1575}}{{{x^2}}} + 14\)

setting their derivative equal to 0 (seen anywhere)     (M1)

e.g. \(\frac{{ - 3150}}{{{x^2}}} + 14 = 0\)

minimum cost is 420 (dollars)     A1     N4

METHOD 2

correct expression for second side, using area = 525     (A1)

e.g. let \({\rm{AD}} = x\) , \({\rm{AB}} = \frac{{525}}{x}\)

attempt to set up cost function using \(\$ 3\) for three sides and \(\$ 11\) for one side     (M1)

e.g. \(3({\rm{AD}} + {\rm{BC}} + {\rm{CD}}) + 11{\rm{AB}}\)

correct expression for cost     A2

e.g. \(3\left( {x + x + \frac{{525}}{x}} \right) + \frac{{525}}{x} \times 11\) , \(3\left( {{\rm{AD}} + {\rm{AD}} + \frac{{525}}{{{\rm{AD}}}}} \right) + \frac{{525}}{{{\rm{AD}}}} \times 11\) , \(6x + \frac{{7350}}{x}\)

EITHER

sketch of cost function     (M1)

identifying minimum point     (A1)

e.g. marking point on graph, \(x = 35\)

minimum cost is 420 (dollars)     A1     N4

OR

correct derivative (may be seen in equation below)     (A1)

e.g. \(C'(x) = 6 - \frac{{7350}}{{{x^2}}}\)

setting their derivative equal to 0 (seen anywhere)     (M1)

e.g. \(6 - \frac{{7350}}{{{x^2}}} = 0\)

minimum cost is 420 (dollars)     A1     N4

[7 marks]

Although this question was a rather straight-forward optimisation question, the lack of structure caused many candidates difficulty. Some were able to calculate cost values but were unable to create an algebraic cost function. Those who were able to create a cost function in two variables often could not use the area relationship to obtain a function in a single variable and so could make no further progress. Of those few who created a correct cost function, most set the derivative to zero to find that the minimum cost occurred at \(x = 15\) , leading to \(\$ 420\). Although this is a correct approach earning full marks, candidates seem not to recognise that the result can be obtained from the GDC, without the use of calculus.