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