Engineers need to lay pipes to connect two cities A and B that are separated by a river of width 450 metres as shown in the following diagram. They plan to lay the pipes under the river from A to X and then under the ground from X to B. The cost of laying the pipes under the river is five times the cost of laying the pipes under the ground.

Let \({\text{EX}} = x\).

 

Let k be the cost, in dollars per metre, of laying the pipes under the ground.

(a)     Show that the total cost C, in dollars, of laying the pipes from A to B is given by \(C = 5k\sqrt {202\,500 + {x^2}}  + (1000 - x)k\).

(b)     (i)     Find \(\frac{{{\text{d}}C}}{{{\text{d}}x}}\).

          (ii)     Hence find the value of x for which the total cost is a minimum, justifying that this value is a minimum.

(c)     Find the minimum total cost in terms of k.

The angle at which the pipes are joined is \({\rm{A\hat XB}} = \theta \).

(d)     Find \(\theta \) for the value of x calculated in (b).

For safety reasons \(\theta \) must be at least 120°.

Given this new requirement,

(e)     (i)     find the new value of x which minimises the total cost;

          (ii)     find the percentage increase in the minimum total cost.

(a)     \(C = {\text{AX}} \times 5k + {\text{XB}} \times k\)     (M1)

 

Note:     Award (M1) for attempting to express the cost in terms of AX, XB and k.

 

\( = 5k\sqrt {{{450}^2} + {x^2}}  + (1000 - x)k\)     A1

\( = 5k\sqrt {202\,500 + {x^2}}  + (1000 - x)k\)     AG

[2 marks]

 

(b)     (i)     \(\frac{{{\text{d}}C}}{{{\text{d}}x}} = k\left[ {\frac{{5 \times 2x}}{{2\sqrt {202\,500 + {x^2}} }} - 1} \right] = k\left( {\frac{{5x}}{{\sqrt {202\,500 + {x^2}} }} - 1} \right)\)     M1A1

 

Note:     Award M1 for an attempt to differentiate and A1 for the correct derivative.

 

          (ii)     attempting to solve \(\frac{{{\text{d}}C}}{{{\text{d}}x}} = 0\)     M1

          \(\frac{5}{{\sqrt {202\,500 + {x^2}} }} = 1\)     (A1)

          \(x = 91.9{\text{ (m) }}\left( { = \frac{{75\sqrt 6 }}{2}{\text{ (m)}}} \right)\)     A1

          METHOD 1

          for example,

          at \(x = 91\frac{{{\text{d}}C}}{{{\text{d}}x}} =  - 0.00895k < 0\)     M1

          at \(x = 92\frac{{{\text{d}}C}}{{{\text{d}}x}} =  0.001506k > 0\)     A1

 

Note:     Award M1 for attempting to find the gradient either side of \(x = 91.9\) and A1 for two correct values.

 

          thus \(x = 91.9\) gives a minimum     AG

          METHOD 2

          \(\frac{{{{\text{d}}^2}C}}{{{\text{d}}{x^2}}} = \frac{{1\,012\,500k}}{{{{\left( {{x^2} + 202\,500} \right)}^{\frac{3}{2}}}}}\)

          at \(x = 91.9\frac{{{{\text{d}}^2}C}}{{{\text{d}}{x^2}}} = 0.010451k > 0\)     (M1)A1

 

Note:     Award M1 for attempting to find the second derivative and A1 for the correct value.

 

Note:     If \(\frac{{{{\text{d}}^2}C}}{{{\text{d}}{x^2}}}\) is obtained and its value at \(x = 91.9\) is not calculated, award (M1)A1 for correct reasoning eg, both numerator and denominator are positive at \(x = 91.9\).

 

          thus \(x = 91.9\) gives a minimum     AG

          METHOD 3

          Sketching the graph of either C versus x or \(\frac{{{\text{d}}C}}{{{\text{d}}x}}\) versus x.     M1

          Clearly indicating that \(x = 91.9\) gives the minimum on their graph.     A1

[7 marks]

 

(c)     \({C_{\min }} = 3205k\)     A1

 

Note:     Accept 3200k.

     Accept 3204k.

 

[1 mark]

 

(d)     \(\arctan \left( {\frac{{450}}{{91.855865{\text{K}}}}} \right) = 78.463{\text{K}}^\circ \)     M1

\(180 - 78.463{\text{K = 101.537K}}\)

\( = 102^\circ \)     A1

[2 marks]

 

(e)     (i)     when \(\theta  = 120^\circ ,{\text{ }}x = 260{\text{ (m) }}\left( {\frac{{450}}{{\sqrt 3 }}{\text{ (m)}}} \right)\)     A1

          (ii)     \(\frac{{133.728{\text{K}}}}{{3204.5407685{\text{K}}}} \times 100\% \)     M1

          \( = 4.17{\text{ (% )}}\)     A1

[3 marks]

 

Total [15 marks]

[N/A]