Question

The volume of a rectangular box with a square base remains constant at 500 cm3 as the area of the base increases at a rate of 10 cm2/sec. Find the rate at which the height of the box is decreasing when each side of the base is 19 cm long. (Do not round your answer.)

Answer 1

Answer:

The rate of change of the height of the box at which is decreasing is $\frac{5000}{130321}$ centimeters per second.

Step-by-step explanation:

From Geometry the volume of a rectangular box ($V$), measured in cubic centimeters, with a square base is modelled by the following formula:

$V = A_{b}\cdot h$ (Eq. 1)

Where:

$A_{b}$ - Area of the base, measured in square centimeters.

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

The height of the box is cleared within the formula:

$h = \frac{V}{A_{b}}$

If we know that $V = 500\,cm^{3}$ and $A_{b} = 361\,cm^{2}$, then the current height of the box is:

$h = \frac{500\,cm^{3}}{361\,cm^{2}}$

$h = \frac{500}{361}\,cm$

The rate of change of volume in time ($\frac{dV}{dt}$), measured in cubic centimeters per second, is derived from (Eq. 1):

$\frac{dV}{dt} = \frac{dA_{b}}{dt}\cdot h + A_{b}\cdot \frac{dh}{dt}$ (Eq. 2)

Where:

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

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

If we get that $\frac{dV}{dt} = 0\,\frac{cm^{3}}{s}$, $\frac{dA_{s}}{dt} = 10\,\frac{cm^{2}}{s}$, $h = \frac{500}{361}\,cm$ and $A_{b} = 361\,cm^{2}$, then the equation above is reduced into this form:

$0\,\frac{cm^{3}}{s} = \left(10\,\frac{cm^{2}}{s} \right)\cdot \left(\frac{500}{361}\,cm \right)+(361\,cm^{2})\cdot \frac{dh}{dt}$

Then, the rate of change of the height of the box at which is decreasing is:

$\frac{dh}{dt} = -\frac{5000}{130321}\,\frac{cm}{s}$

The rate of change of the height of the box at which is decreasing is $\frac{5000}{130321}$ centimeters per second.