The side of the base of a square prism is decreasing at a rate of 7 km/min and the height of the prism is increasing at a rate of 10 km/min. At a certain instant, the base side is 4 km and the height is 9 km. What is the rate of change of the surface area of the prism at that instant

  -204 km^2/min

Step-by-step explanation:

For base edge length s and height h, the surface area of the prism is ...

  A = 2(s^2 +2sh) = 2s^2 +4sh

Then the rate of change of surface area is ...

  A' = 4s·s' +4s'·h +4s·h' = 4s(s' +h') +4s'·h

Filling in the given values, we find the rate of change of area to be ...

  A' = 4(4 km)(-7 km/min +10 km/min) +4(-7 km/min)(9 km)

  A' = 4(12 km^2/min -63 km^2/min)

  A' = -204 km^2/min

The area is decreasing at the rate of 204 square km per minute.

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