QCE Vault / Mathematical Methods Further applications of differentiation — Question 88 Original QCE Vault practice · 5 marks
Browse all questions Differentiation revision Original practice exam Report an issue Q88 · Practice question Technology-active Complex familiar 5 marks
QUESTION 88 (5 marks) A closed cylindrical can has volume 500 π cm 3 500\pi\text{ cm}^3 500 π cm 3 . Let its radius be r r r cm and height be h h h cm. Determine the radius and height that minimise its total surface area. WORKED SOLUTION
Practice marking scheme 5 marks ANSWER r ≈ 6.30 cm r\approx6.30\text{ cm} r ≈ 6.30 cm and h ≈ 12.6 cm h\approx12.6\text{ cm} h ≈ 12.6 cm . Worked solution
From π r 2 h = 500 π \pi r^2h=500\pi π r 2 h = 500 π , h = 500 / r 2 h=500/r^2 h = 500/ r 2 . Thus S = 2 π r 2 + 2 π r h = 2 π r 2 + 1000 π / r S=2\pi r^2+2\pi rh=2\pi r^2+1000\pi/r S = 2 π r 2 + 2 π r h = 2 π r 2 + 1000 π / r . Setting S ′ ( r ) = 4 π r − 1000 π / r 2 = 0 S'(r)=4\pi r-1000\pi/r^2=0 S ′ ( r ) = 4 π r − 1000 π / r 2 = 0 gives r 3 = 250 r^3=250 r 3 = 250 , so r ≈ 6.30 r\approx6.30 r ≈ 6.30 and h = 500 / r 2 ≈ 12.6 h=500/r^2\approx12.6 h = 500/ r 2 ≈ 12.6 . Uses the volume constraint to obtain h = 500 / r 2 h=500/r^2 h = 500/ r 2 . [1 mark] Forms S ( r ) = 2 π r 2 + 1000 π / r S(r)=2\pi r^2+1000\pi/r S ( r ) = 2 π r 2 + 1000 π / r . [1 mark] Differentiates and sets the derivative equal to zero.
[1 mark] Obtains r ≈ 6.30 r\approx6.30 r ≈ 6.30 cm. [1 mark] Obtains h ≈ 12.6 h\approx12.6 h ≈ 12.6 cm and identifies the minimum. [1 mark] Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabus How many marks did you earn? Compare your working with the guide above.
