What is the rate of change of the height, in cm/min when the height is in 1cm?

Question

A rectangular box has a width of 2cm and a length of 5cm. The volume of the box is decreasing at a rate of 4cm3/min, with the width and the length being held constant. What is the rate of change of the height, in cm/min, when the height is 1cm?

Submit an exact answer in fractional form.

Answer 1

What is the rate of change of the height, in cm/min when the height is in 1cm? A rectangular box has a width of 2cm and a length of 5cm. The volume of the box is decreasing at a rate of 4cm³/min, with the width and the length being held constant. What is the rate of change of the height, in cm/min, when the height is 1cm? Submit an exact answer in fractional form.

Detailed Solution:

Given: w = 2 cm l = 5 cm h = 1 cm dV/dt = -4cm³/min dh/dt = ?? w & l are constant

Volume of Box = Area of Base * Height ==> V = l * w * h ==> V = 5 * 2 * h (w & l are constant)

V = 10h ==> d/dt (V = 10h) ==> dV/dt = 10dh/dt ==> 10dh/dt = dV/dt ==> dh/dt = (dV/dt)/(10) ==>

dh/dt = (dV/dt)/(10) = (-4)/(10) = -2/5 cm/min (Final Answer)

While the width and length are held constant, the volume of the box is decreasing at a rate of 4cm³/min (dV/dt = -4cm³/min), the height of the box is decreasing at a rate of 2/5 cm/min (dh/dt = -2/5 cm/min)

Answer 2

Natasha,

Hello again,

Vol = (2)(5)(h)

take derivative of both sides and plug in the values

Hope this helps--Andrew F.