Answer:
The area of the rectangle is increasing at a rate of 169 cm²/s
Step-by-step explanation:
Given;
increase in the length of the rectangle, [tex]\frac{dL}{dt} = 7 \ cm/s[/tex]
increase in the width of the rectangle, [tex]\frac{dW}{dt} = 8 \ cm/s[/tex]
length, L = 15 cm
width, W = 7 cm
The increase in Area is calculated as;
[tex]Area = Length \times Width\\\\A = LW\\\\\frac{dA}{dt} = L(\frac{dW}{dt} )\ + \ W(\frac{dL}{dt} )\\\\\frac{dA}{dt} = 15 \ cm(8\ \frac{ cm}{s} ) \ + \ 7 \ cm(7\ \frac{ cm}{s} ) \\\\\frac{dA}{dt} = 120 \ cm^2/s \ + \ 49 \ cm^2/s\\\\\frac{dA}{dt} = 169 \ cm^2/s[/tex]
Therefore, the area of the rectangle is increasing at a rate of 169 cm²/s
