Ask Question
26 September, 05:28

A rectangular box has a width of 3cm and a length of 2cm. The volume of the box is decreasing at a rate of 2 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 5cm

+4
Answers (1)
  1. 26 September, 07:26
    0
    dh/dt = 0,111 cm/min

    Step-by-step explanation:

    Volume of the rectangular box is:

    V = V (b) = Area of the rectanglar base * height

    Area of the base is : 3*2 = 6 cm²

    V (b) = 6*h

    Differentiating in relation to time in both sides of the equation give:

    d (V (b)) / dt = 6 * dh/dt (1)

    According to problem statement volume of the box is decreasing at the rate of 2 cm each 3 min then by rule of three

    2 cm ⇒ 3 min

    x ⇒ 1 min

    x = 2/3 = 0,67 cm/min

    As dimensions of the box (length and width) will be constants decreasing in the volume of the cube does not depend on the level of the height, and we know dV (b) / dt = 0.67 cm/min, then in equation 1

    0.67 = 6*dh/dt

    dh/dt = 0.67/6

    dh/dt = 0,111 cm/min
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A rectangular box has a width of 3cm and a length of 2cm. The volume of the box is decreasing at a rate of 2 cm 3/min, with the width and ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers