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Question

The radius of the base of a cone is increasing at the rate of 3 cm/min and the altitude is decreasing at the rate of 4 cm/min. The rate of change of lateral surface area when the radius is 7 cm and the altitude is 24 cm is


A

54π cm2/min

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B

7π cm2/min

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C

27π cm2/min

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D

108π cm2/min

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Solution

The correct option is A

54π cm2/min


Let r, l, and h denote, respectively, the radius, slant height, and altitude of the cone at any time t.

Then, l2=r2+h2

Differentiating, we get

2ldldt=2rdrdt+2hdhdt

=ldldt=rdrdt+hdhdt

ldldt=7×3+24×(4) [dhdt=4 and drdt=3]

ldldt=75

Also l2=72+242 [r=7 and h=24]

l=625=25 cm

dldt=7525=3 cm/min

Let S denote the lateral surface area. Then

S=πrl

dSdt=ddt(πrl)= π(drdtl+rdldt)

=π(3×25+7×(3))

=54π cm2/min


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