A rectangle's base remains 0.5 cm while its height changes at a rate of 1.5 cm/min. At what rate is the area changing, in cm when the height is 1.5 cm?

Question

William Abdalla · Accepted Answer

The area of the rectangle is changing at #0.75 # **sq.cm/min.**

Let #x and y # be the base and height of the rectangle. #x=0.5# cm is constant :. dx/dt=0#. Area of the rectangle is #A=x*y# . Differentiating both sides we get #(dA)/dt= dx/dt*y +dy/dt*x ;y=1.5, dy/dt=1.5# cm /min. #:.(dA)/dt= 0*1.5 +1.5*0.5= 0.75 # sq.cm/min. The area of the rectangle is changing at #0.75 # sq.cm/min.[Ans]

Scarlett Allison · Answer

undefined To find the rate at which the area is changing when the height is 1.5 cm, we use the formula for the area of a rectangle, which is given by length multiplied by width (base multiplied by height).

Given:
Base (b) = 0.5 cm
Height (h) = 1.5 cm
Rate of change of height (dh/dt) = 1.5 cm/min

We are asked to find the rate of change of area (dA/dt) when the height is 1.5 cm.

Using the formula for the area of a rectangle:

A = b * h

We differentiate both sides of the equation with respect to time (t):

dA/dt = (db/dt) * h + b * (dh/dt)

Since the base remains constant, db/dt = 0.

Substituting the given values:

dA/dt = 0 * 1.5 + 0.5 * 1.5

Solving this:

dA/dt = 0 + 0.75

Therefore, when the height is 1.5 cm, the rate at which the area is changing is 0.75 cm²/min.