The question is related to application of derivatives .
Let A represents the area and s represent the side length .
[tex] A = s^2 [/tex]
[tex] 25=s^2 [/tex]
[tex] s =\sqrt{25} = 5 [/tex]
[tex] A = s^2 [/tex]
Differentiating both sides with respect to t
It is given that ds/dt = 4cm/s
Substituting the values of s, ds/dt and on multiplying them, we will get
dA/dt = 2(5)(4) = 40 cm^2/s
Therefore rate of change of area when area=[tex] 25 cm^2 [/tex] and rate of change of side is 4 cm/s is 40 [tex] cm^2/s . [/tex]
