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.