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10. Prof. Feinman And Her Husband Decided To Taste A Frozen Durian, A Tropical Fruit With A Unique Strong Aroma. Each Piece Of Durian They Bought Came In The Shape Of A Perfect Cube. The Side Of The First Durian Cube Prof. Feinman Tried Decreased At 2 Cm Per Minute. At What Rate Was The Durian's Surface Area Changing When The Side Of The Durian Was 4 Cm ?

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Let's start by identifying the relevant formula for the surface area of a cube. The surface area (A) of a cube with side length (s) is:

A = 6s^2

We are given that the side length of the cube is changing at a rate of -2 cm/min (the negative sign indicates that the side length is decreasing). We want to find the rate of change of the surface area when the side length is 4 cm.

To solve this problem, we can use the chain rule of differentiation. We have:

dA/dt = dA/ds * ds/dt

where dA/dt is the rate of change of the surface area, dA/ds is the rate of change of the surface area with respect to the side length (which we can find by differentiating the surface area formula), ds/dt is the rate of change of the side length, and t is time.

Differentiating the surface area formula with respect to the side length, we get:

dA/ds = 12s

Plugging in s = 4 cm (since we want to find the rate of change when the side length is 4 cm), we get:

dA/ds = 12(4) = 48 cm^2

We are given that ds/dt = -2 cm/min (since the side length is decreasing at a rate of 2 cm per minute). Plugging in these values, we get:

dA/dt = (48 cm^2/cm) * (-2 cm/min) = -96 cm^2/min

Therefore, the rate of change of the surface area when the side of the durian is 4 cm is -96 cm^2/min. Note that the negative sign indicates that the surface area is decreasing at a rate of 96 cm^2 per minute.
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