Answer to Question #127914 in Calculus for alex

Question

A closed cylindrical tin is of height h cm and radius r cm, its total surface area is A cm2 and its volume is r cm3. Find an expression for A in terms of r . Taking , find an expression of v in terms of r , hence determine the value of r which make v maximum.

Answer 1

Radius of cylinder = r cm

Height of cylinder = h cm

Volume of cylinder = πr²h cm³

But given that volume= r cm³

So πr²h = r

=> h = r/πr²

=> h = 1/πr

=> πrh = 1

Total surface area = 2πrh + 2πr²

So A = 2πrh + 2πr² = 2+ 2πr²

=> A = 2+ 2πr²

This is the expression for A in term of r

When A is constant,

A = 2πrh + 2πr²

=> 2πrh = A - 2πr²

So volume v = πr²h =$\frac{ r(A-2πr²)}{2}$

=> v = $\frac{Ar}{2}$ - πr³

$So \frac {dv}{dr}$ = $\frac {A}{2}$ - 3πr²

For maximum value of v, $\frac {dv}{dr} =0$

=> $\frac {A}{2}$ - 3πr² = 0

=> r² = $\frac {A}{6π}$

=> r = $\sqrt{\frac {A}{6π}}$

$\frac {d²v}{dr²}$ = -6πr

Obviously $\frac {d²v}{dr²}$ is negative when r = $\sqrt{\frac {A}{6π}}$

So v is maximum when r = $\sqrt{\frac {A}{6π}}$