Question

A box currently has length and width of 10 inches and a height of 3 inches. Use calculus to determine at what rate the height changes, if the length and width are decreasing at a rate of 2 inches per minute and the volume of the box is constant. (6 points)

Answer 1

Volume of Box = length * width * height

length = width = 10 inch

height = 3 inch

d/dt(length) = d/dt(width) = -2 inch/min

d/dt(Volume of Box) = length * width * d/dt(height) + length * d/dt(width) * height + d/dt(length) * width * height

0 = 10(10)(rate of change of height) + 10(-2)(3) + (-2)(10)(3)

0 = 100(rate of change of height) - 60 - 60

rate of change of height = 120/100 = 1.2 inches per minute

Hence height is increasing at the rate of 1.2 inches per minute to keep the volume of box as constant

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