A rectangular box has a width of 7 cm and a length of 1 cm . The volume of the box is decreasing at a rate of 3 cm 3 / 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 7 cm ?



Answer :

Answer: [tex]\frac{3}{7} \text{ cm/min}[/tex]

Step-by-step explanation:

The volume of the box is [tex]V=lwh=7h[/tex].

Differentiating both sides with respect to [tex]t[/tex] yields [tex]\frac{dV}{dt}=7 \frac{dh}{dt}[/tex].

It is given that [tex]\frac{dV}{dt}=-3[/tex], which yields [tex]\frac{dh}{dt}=-\frac{3}{7}[/tex].

Therefore, the height of the box is decreasing at a rate of [tex]\frac{3}{7} \text{ cm/min}[/tex].