We know that a soap bubble will be in sphere shape
Consider the radius of soap bubble as \( r \dfrac{dr}{dt}\) = 0.2 cm/s
The surface area of soap bubble = 4πr2
So the rate of change of surface area = 8πr
By substituting the values = 8 × 3.14 × 7 × 0.2
We get,
\( \dfrac{dS}{dt}\)= 35.2 cm2/s
Answered by Sakshi | 1 year agoGiven the sum of the perimeters of a square and a circle, show that the sum of their areas is least when one side of the square is equal to diameter of the circle.
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