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odolwa99
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I have managed to solve (i), so I'll just post the answer as it comes into play for (ii). I'm struggling with the differentiation of this expression. Can anyone help?
Many thanks.
Q. A tank, with a base, is made from a thin uniform metal. The tank, standing on level ground, is in the shape of an upright circular cylinder & hemispherical top, with radius length of r metres. The height of the cylinder is h metres. (i) If the total surface area of the tank is 45∏m2, express h in terms of r, (ii) Find the values of h & r, for which the tank has maximum volume.
Attempt: (i) [itex]\frac{45 - 3r^2}{2r}[/itex]
(ii) 1st, separate the fractions and simplify the answer in (i) to [itex]\frac{45}{2r}[/itex] - [itex]\frac{3r}{2}[/itex]
[itex]\frac{dS}{dx}[/itex] = -[itex]\frac{45}{r^2}[/itex] - [itex]\frac{3}{2}[/itex] = 0 => [itex]\frac{3r^2}{2}[/itex] = -45 => 3r2 = -90 => r2 = -30 => r = -[itex]\sqrt{30}[/itex]
Ans: (From textbook): r = 3
Many thanks.
Homework Statement
Q. A tank, with a base, is made from a thin uniform metal. The tank, standing on level ground, is in the shape of an upright circular cylinder & hemispherical top, with radius length of r metres. The height of the cylinder is h metres. (i) If the total surface area of the tank is 45∏m2, express h in terms of r, (ii) Find the values of h & r, for which the tank has maximum volume.
Homework Equations
The Attempt at a Solution
Attempt: (i) [itex]\frac{45 - 3r^2}{2r}[/itex]
(ii) 1st, separate the fractions and simplify the answer in (i) to [itex]\frac{45}{2r}[/itex] - [itex]\frac{3r}{2}[/itex]
[itex]\frac{dS}{dx}[/itex] = -[itex]\frac{45}{r^2}[/itex] - [itex]\frac{3}{2}[/itex] = 0 => [itex]\frac{3r^2}{2}[/itex] = -45 => 3r2 = -90 => r2 = -30 => r = -[itex]\sqrt{30}[/itex]
Ans: (From textbook): r = 3