Given:
length of a rectangle is increasing at a rate of 5 in. /s and width is decreasing at 3 in. /s
Let x and y are the length and width of the rectangle respectively.
[tex]\frac{dx}{dt}=5\text{ in./s ; }\frac{dy}{dt}=-3\text{ in./s ; x=}45\text{ in. ; y=35 in.}[/tex]
Area of the rectangle (A)= length(x) X width(y)
[tex]A=x\times y[/tex]
Differentiate with respect to t
[tex]\frac{dA}{dt}=\text{x}\frac{\text{dy}}{dt}+\text{y}\frac{\text{dx}}{dt}[/tex][tex]\frac{dA}{dt}=45(-3)+35(5)[/tex][tex]\frac{dA}{dt}=-135+175[/tex][tex]\frac{dA}{dt}=40in^2\text{ /s}[/tex][tex]\text{Rate of change of its area is 40 in}^2\text{ /s}[/tex]
