If \(\alpha \) is the angle between [CD] and the wall, show that \(L = \frac{a }{{\sin \alpha }} + \frac{b}{{\cos \alpha }}{\text{, }}0 < \alpha  < \frac{\pi }{2}\).

If a = 5 and b = 1, find the maximum length of a painting that can be removed through this doorway.

Let a = 3k and b = k .

Find \(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }}\).

Let a = 3k and b = k . 

Find, in terms of k , the maximum length of a painting that can be removed from the gallery through this doorway.

Let a = 3k and b = k . 

Find the minimum value of k if a painting 8 metres long is to be removed through this doorway.

\(L = {\text{CA}} + {\text{AD}}\)     M1

\({\text{sin}}\alpha {\text{ = }}\frac{a}{{{\text{CA}}}} \Rightarrow {\text{CA}} = \frac{a}{{\sin \alpha }}\)     A1

\(\cos \alpha  = \frac{b}{{{\text{AD}}}} \Rightarrow {\text{AD}} = \frac{b}{{\cos \alpha }}\)     A1

\(L = \frac{a}{{\sin \alpha }} + \frac{b}{{\cos \alpha }}\)     AG

[2 marks]

\(a = 5{\text{ and }}b = 1 \Rightarrow L = \frac{5}{{\sin \alpha }} + \frac{1}{{\cos \alpha }}\)

METHOD 1

     (M1)

minimum from graph \( \Rightarrow L = 7.77\)     (M1)A1

minimum of L gives the max length of the painting     R1

[4 marks]

METHOD 2

\(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = \frac{{ - 5\cos \alpha }}{{{{\sin }^2}\alpha }} + \frac{{\sin \alpha }}{{{{\cos }^2}\alpha }}\)     (M1)

\(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = 0 \Rightarrow \frac{{{{\sin }^3}\alpha }}{{{{\cos }^3}\alpha }} = 5 \Rightarrow \tan \alpha  = \sqrt[{3{\text{ }}}]{5}{\text{ }}(\alpha  = 1.0416...)\)     (M1)

minimum of L gives the max length of the painting     R1

maximum length = 7.77     A1

[4 marks]

\(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = \frac{{ - 3k\cos \alpha }}{{{{\sin }^2}\alpha }} + \frac{{k\sin \alpha }}{{{{\cos }^2}\alpha }}\,\,\,\,\,{\text{(or equivalent)}}\)     M1A1A1

[3 marks]

\(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = \frac{{ - 3k{{\cos }^3}\alpha  + k{{\sin }^3}\alpha }}{{{{\sin }^2}\alpha {{\cos }^2}\alpha }}\)     (A1)

\(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = 0 \Rightarrow \frac{{{{\sin }^3}\alpha }}{{{{\cos }^3}\alpha }} = \frac{{3k}}{k} \Rightarrow \tan \alpha  = \sqrt[3]{3}\,\,\,\,\,(\alpha  = 0.96454...)\)     M1A1

\(\tan \alpha  = \sqrt[3]{3} \Rightarrow \frac{1}{{\cos \alpha }} = \sqrt {1 + \sqrt[3]{9}} \,\,\,\,\,(1.755...)\)     (A1)

\({\text{and }}\frac{1}{{\sin \alpha }} = \frac{{\sqrt {1 + \sqrt[3]{9}} }}{{\sqrt[3]{3}}}\,\,\,\,\,(1.216...)\)     (A1)

\(L = 3k\left( {\frac{{\sqrt {1 + \sqrt[3]{9}} }}{{\sqrt[3]{3}}}} \right) + k\sqrt {1 + \sqrt[3]{9}} \,\,\,\,\,(L = 5.405598...k)\)     A1     N4

[6 marks]

\(L \leqslant 8 \Rightarrow k \geqslant 1.48\)     M1A1

the minimum value is 1.48

[2 marks]

Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC. 

In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for \(\alpha \) and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question.

Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC. 

In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for \(\alpha \) and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question.

Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC. 

In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for \(\alpha \) and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question.

Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC. 

In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for \(\alpha \) and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question.

Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC. 

In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for \(\alpha \) and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question.