The volume, [tex]V \, \text{cm}^3[/tex], of a tin of radius [tex]r \, \text{cm}[/tex] is given by the formula

\[V = \pi (40r - r^2 - r^3)\]

Find the positive value of [tex]r[/tex] for which

\[\frac{dV}{dr} = 0\]

and find the value of [tex]V[/tex] which corresponds to this value of [tex]r[/tex].

V = π ( 40 r − r 2 − r 3 ) V = 40 π r − π r 2 − π r 3 d r d V ​ = 40 π − 2 π r − 3 π r 2 d r d V ​ = 0 40 π − 2 π r − 3 π r 2 = 0 40 − 2 r − 3 r 2 = 0 3 r 2 + 2 r − 40 = 0 3 r 2 + 12 r − 10 r − 40 = 0 3 r ( r + 4 ) − 10 ( r + 4 ) = 0 ( 3 r − 10 ) ( r + 4 ) r = − 4 , 10/3
r is positive, so **r =10/3 cm **
So, if r =10/3: **
V = π ( 40 r − r 2 − r 3 ) V = π ( 40 ( 10/3 ) − ( 10/3 ) 2 − ( 10/3 ) 3 ) V = π ( 400/3 − 100/9 − 1000/27 ) V = π ( 85.185... ) V = 85.185... π V = 267.617... V = 268 c m 3 ( 3 s . f . )
**

Answered by ollieboyne | 2024-06-10

The positive value of r for which d r d V ​ = 0 is 3 10 ​ cm. The corresponding volume V at this radius is approximately 268 cm³. This indicates the volume of the tin at the optimized radius.
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Answered by ollieboyne | 2024-09-26