Question

The volume of a rectangular box with a square base remains constant at 1000 cm3 as the area of the base increases at a rate of 4 cm2 /sec. Find the rate at which the height of the box is decreasing when each side of the base is 12 cm long.

Answer 1

Answer:

The rate at which the height of the box is decreasing when each side of the base is 12 centimeters long is 0.193 centimeters per second.

Step-by-step explanation:

We can represent the volume of the rectangular box with a square base ($V$), measured in cubic centimeters, by means of this formula:

$V = A\cdot h$ (Eq. 1)

Where:

$A$ - Area of the base of the box, measured in centimeters.

$h$ - Height of the rectangular box, measured in centimeters.

Given that volume remains constant, we clear the height of the box as follows:

$h = V\cdot A^{-1}$ (Eq. 1b)

By Differential Calculus, we get the expression for the rate of change of the height:

$\frac{dh}{dt} = - V\cdot A^{-2}\cdot \frac{dA}{dt}$ (Eq. 2)

Where:

$\frac{dA}{dt}$ - Rate of change of the area of the base in time, measured in square centimeters per second.

$\frac{dh}{dt}$ - Rate of change of the height of the box in time, measured in centimeters per second.

If we know that $V = 1000\,cm^{3}$, $A = 144\,cm^{2}$ and $\frac{dA}{dt} = 4\,\frac{cm^{2}}{s}$, then the rate of change of the area of the base in time is:

$\frac{dh}{dt} = -(1000\,cm^{3})\cdot (144\,cm^{2})^{-2} \cdot \left(4\,\frac{cm^{2}}{s} \right)$

$\frac{dh}{dt} = -0.193\,\frac{cm}{s}$

The rate at which the height of the box is decreasing when each side of the base is 12 centimeters long is 0.193 centimeters per second.