The radius of the base of a cone is increasing at the rate of 3 cm/minute and the altitude is decreasing at the rate of 4 cm/minute. At what rate, lateral surface is changing when the radius in 7 cm and altitude is 24 cm?

Options

A. `54pi cm^(2)//min`
B. `7 pi cm^(2)//min`
C. `27 cm^(2)//min`
D. none of these

Answer

A

Explanation

Let r, l and h denote respectively the radius, slant height and height of the cone at any time l. Then,
`l^(2)=r^(2)+h^(2)`
`implies 2l(dl)/(dt)=2r(dr)/(dt)+2h(dh)/(dt)`
`implies l(dl)/(dt)=r(dr)/(dt)+h(dh)/(dt)`
`l(dl)/(dt)=7xx3+24xx-4 " "[because(dh)/(dt)=-4and (dr)/(dt)=3]`
`implies l(dl)/(dt)=-75`
When r = 7 and h=24, we have
`l^(2)=7^(2)+24^(2)" "[because l^(2)=r^(2)+h^(2)]`
implies l = 25
`l(dl)/(dt)=-75implies(dl)/(dt)=-3`
Let S denote the laternal surface area. Then,
`S=pi r l`
`implies (dS)/(dt)=pi{(dr)/(dt)l+r(dl)/(dt)}=pi{3xx25+7xx-3}=54pi`