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A rectangle has height and width changing in such a way that the area remains constant 2 square feet. At the instant the height is 2 feet, the height is changing at a rate of 6 feet/minute. Find the rate of change of the width (in feet/min) at that instant.

Answer :

Using implicit differentiation, it is found that the rate of change is of the width is of -3 feet/min.

What is the area of a rectangle?

The area of a rectangle of length l and height h is given by:

[tex]A = lh[/tex]

Applying implicit differentiation, the rate of change is given by:

[tex]\frac{dA}{dt} = l\frac{dh}{dt} + h\frac{dl}{dt}[/tex]

In this problem, we have that:

  • The area is constant, hence [tex]\frac{dA}{dt} = 0[/tex].
  • Area of 2 square feet, height of 2 feet, hence [tex]h = 2, lh = 2 \rightarrow l = 1[/tex].
  • The height is changing at a rate of 6 feet/minute, hence [tex]\frac{dh}{dt} = 6[/tex].

Then:

[tex]\frac{dA}{dt} = l\frac{dh}{dt} + h\frac{dl}{dt}[/tex]

[tex]0 = 6 + 2\frac{dl}{dt}[/tex]

[tex]2\frac{dl}{dt} = -6[/tex]

[tex]\frac{dl}{dt} = -3[/tex]

The rate of change is of the width is of -3 feet/min.

To learn more about implicit differentiation, you can take a look at https://brainly.com/question/25608353