Class 12 Applications Of Derivatives · Case Study4 Markscase_based_application
Class 12 Maths · Applications Of Derivatives · Case Study #29
An open-top right circular cylinder is formed from a metal sheet of area 75π cm², with negligible thickness and no waste. Its radius and height are r,h>0; material forms the base and curved wall, so πr²+2πrh=75π. The engine-bore context is illustrative; use this open-cylinder model.
Part (1) 1 Marks
Express its volume V as a function of r and state the feasible domain.
Answer: V=(π/2)(75r−r³), 0<r<√75
- The material constraint gives h=(75−r²)/(2r); positivity gives 0<r<√75.
- Substitute in V=πr²h to get V=(π/2)(75r−r³).
Part (2) 1 Marks
Find dV/dr.
Answer: (3π/2)(25−r²)
- Differentiate V=(π/2)(75r−r³).
- dV/dr=(π/2)(75−3r²)=(3π/2)(25−r²).
Part (3) 2 Marks
Find the radius for maximum volume.
Answer: 5 cm
- V′=(3π/2)(25−r²)=0 gives r=5 in the feasible domain.
- V′>0 for 0<r<5 and V′<0 for 5<r<√75, proving the global maximum.
- The maximizing radius is 5 cm.
Part (4) 2 Marks
Alternatively, is h>r at maximum volume? Justify.
Answer: False: h=r=5 cm.
- V′=(3π/2)(25−r²) changes from positive to negative at r=5, the maximum.
- Substitute into h=(75−r²)/(2r): h=(75−25)/10=5.
- Thus h=r=5 cm, so h>r is false.