the width of a rectangle is increasing at a rate of 5 inches per second and its length is increasing at the…

the width of a rectangle is increasing at a rate of 5 inches per second and its length is increasing at the rate of 3 inches per second. at what rate is the area of the rectangle increasing when its width is 2 inches and its length is 7 inches? (hint: let (w(t)) and (l(t)) be the width and length, respectively, at time (t).) the rate that the area of the rectangle is increasing is (square) in²/s. (simplify your answer.)

the width of a rectangle is increasing at a rate of 5 inches per second and its length is increasing at the rate of 3 inches per second. at what rate is the area of the rectangle increasing when its width is 2 inches and its length is 7 inches? (hint: let (w(t)) and (l(t)) be the width and length, respectively, at time (t).) the rate that the area of the rectangle is increasing is (square) in²/s. (simplify your answer.)

Answer

Explanation:

Step1: Recall area formula

Let $w$ be the width and $l$ be the length of the rectangle. The area $A = wl$.

Step2: Differentiate with respect to time

Using the product - rule $\frac{dA}{dt}=l\frac{dw}{dt}+w\frac{dl}{dt}$.

Step3: Identify given values

We know that $\frac{dw}{dt}=5$ inches per second, $\frac{dl}{dt}=3$ inches per second, $w = 2$ inches and $l = 7$ inches.

Step4: Substitute values

$\frac{dA}{dt}=(7\times5)+(2\times3)$.

Step5: Calculate result

$\frac{dA}{dt}=35 + 6=41$.

Answer:

$41$