Answer: [tex]30\ in.^2/min[/tex]
Step-by-step explanation:
Given
The side of a square is increasing at the rate of [tex]\frac{dl}{dt}=3\ in./min[/tex]
The area of the square is the square of the side length
[tex]A=l^2[/tex]
when the area is [tex]25\ in.^2[/tex], its side must be [tex]5\ in.[/tex]
Differentiate area w.r.t time
[tex]\Rightarrow \dfrac{dA}{dt}=2l\dfrac{dl}{dt}\\\\\text{Insert the value}\\\Rightarrow \dfrac{dA}{dt}=2\times 5\times 3\\\\\Rightarrow \dfrac{dA}{dt}=30\ in.^2/min[/tex]
