Answer :
The surface area, A, is changing at a rate of 288 km/min decreasing.
Given,
Base length, l = 6 km
Height, h = 10 km
dl/dt = -9 km/min
dh/dt = 12 km/min
Surface area of a prism, A = 2 × (lw + lh + wh)
For a square prism,
Length, l = width, w = 6km
A = 2 × (l² + lh + lh)
= 2(l²) + 4lh
A = 2(l²) + 4 lh
= 2 × (l² + 2lh)
dA/dt = 2 × (2I × dI/dt + 0 × dh/dt + 2 × 2h × dl/dt + 2l × dh/dt)
dA/dt = 2 × (2l × dl/dt + 0 + 2 × 2h × dl/dt + 2l × dh/dt)
= 2 × (2(6) × (-9) + 2(10) × (-9) + 2(6) × (12))
= 2 × (- 108 + -180 + 144)
= 2 × -144
= -288 km/min
The rate of change of the surface area, A is decreasing by 288 km/min.
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Question is incomplete. Completed question is given below;
The length s(t)s(t)s, (, t, )of the side of the base of a square prism is decreasing at a rate of 9 kilometers per minute and the height h(t)h(t)h, (, t, )of the prism is increasing at a rate of 12 kilometers per minute. At a certain instant t_0t 0 t, start subscript, 0, end subscript, the base's side is 6 kilometers and the height is 10 kilometers. What is the rate of change of the surface area A(t)A(t)A, (, t, )of the prism at that instant?
