see the figure below to better understand the problem
The area of the square is
[tex]A=b^2[/tex]
The area of the circle is given by
[tex]A=\pi r^2[/tex]
The area outside the circle but inside the square is given by the equation
[tex]A=b^2-\pi r^2[/tex]
Find out dA/dt
using implicit differentiation
[tex]\frac{dA}{dt}=2b\frac{db}{dt}-2\pi r\frac{dr}{dt}[/tex]
Remember that
we have
db/dt=-5 m/min
dr/dt=-4 m/min
substitute
[tex]\begin{gathered} \frac{dA}{dt}=2b(-5)-2\pi r(-4) \\ \\ \frac{dA}{dt}=-10b+8\pi r \end{gathered}[/tex]
Evaluate for r=2 m and b=24 m
[tex]\begin{gathered} \frac{dA}{dt}=-10(24)+8\pi(2) \\ \\ \frac{dA}{dt}=-240+16\pi \\ \\ \frac{dA}{dt}=-189.73\text{ }\frac{m^2}{min} \end{gathered}[/tex]
Round to the nearest whole number
the answer is -190 m2/min (negative because is decreasing)
