The area of a square is defined by the formula
A = s²
where A is the area and s is the length of one side
you are given with the following data
[tex] \frac{ds}{dt} = 5 m/min[/tex]
[tex]s = 6m[/tex]
since this involves the change of area as to time then having the area differentiated you'll have:
[tex]A = s^2[/tex]
[tex] \frac{dA}{dt} = (2s) \frac{ds}{dt} [/tex]
substituting the given you'll have:
[tex] \frac{dA}{dt} = 2(6m)(5m/min)[/tex]
[tex] \frac{dA}{dt} = 60 m^2/min[/tex]
