SOLUTION 2: Draw a rectangle with length $x$ and width $y$, and assume each edge of the rectangle is a function of time $t$.

$\ \ \ \$ a.) The perimeter of the rectangle is $$P=x+x+y+y \ \ \ \ \longrightarrow$$ $$P = 2x+2y$$ GIVEN: $\ \ \ \displaystyle{ dx \over dt } = 4 \ ft/min. \$ and $\ \displaystyle{ dy \over dt } = -3 \ ft/min.$

FIND: $\ \ \ \displaystyle{ dP \over dt }$ when $x=8 \ ft.$ and $y=5 \ ft.$

Now differentiate the perimeter equation with respect to time $t$ getting

$$D \{ P \} = D \{2x+2y \} \ \ \ \longrightarrow$$ $$\displaystyle{ dP \over dt } = 2 \displaystyle{ dx \over dt } + 2 \displaystyle{ dy \over dt } \ \ \ \longrightarrow$$

$\Big($ Now let $\displaystyle{ dx \over dt } = 4$ and $\ \displaystyle{ dy \over dt } = -3. \Big)$

$$\displaystyle{ dP \over dt } = 2 (4) + 2 (-3 ) = 2 \ ft/min.$$

$\ \ \ \$ b.) The area of the rectangle is $$A= (length)(width) \ \ \ \ \longrightarrow$$ $$A= xy$$ GIVEN: $\ \ \ \displaystyle{ dx \over dt } = 4 \ ft/min.$ and $\displaystyle{ dy \over dt } = - 3 \ ft/min.$

FIND: $\ \ \ \displaystyle{ dA \over dt }$ when $x=8 \ ft.$ and $y=5 \ ft.$

Now differentiate the area equation with repect to time $t$ using the product rule getting

$$D \{ A \} = D \{xy \} \ \ \ \longrightarrow$$ $$\displaystyle{ dA \over dt } = x \displaystyle{ dy \over dt } + \displaystyle{ dx \over dt } y \ \ \ \longrightarrow$$

$\Big($ Now let $\displaystyle{ dx \over dt } = 4$ , $\ \displaystyle{ dy \over dt } = -3 , x=8, \$and $y=5. \Big)$

$$\displaystyle{ dA \over dt } = (8)(-3) + (4)(5) = -4 \ ft^2/min.$$