the width of a rectangle is increasing at a rate of 7 inches per second and its length is increasing at the…

the width of a rectangle is increasing at a rate of 7 inches per second and its length is increasing at the rate of 9 inches per second. at what rate is the area of the rectangle increasing when its width is 2 inches and its length is 8 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 in²/s. (simplify your answer.)

the width of a rectangle is increasing at a rate of 7 inches per second and its length is increasing at the rate of 9 inches per second. at what rate is the area of the rectangle increasing when its width is 2 inches and its length is 8 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 in²/s. (simplify your answer.)

Answer

Explanation:

Step1: Recall area formula

The area of a rectangle is $A = L\times W$, where $L$ is length and $W$ is width.

Step2: Differentiate with respect to time

Using the product - rule $\frac{d(uv)}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}$, we have $\frac{dA}{dt}=L\frac{dW}{dt}+W\frac{dL}{dt}$.

Step3: Substitute given values

We know that $\frac{dW}{dt}=7$ in/s, $\frac{dL}{dt}=9$ in/s, $W = 2$ in and $L = 8$ in. Substitute these values into the formula: $\frac{dA}{dt}=8\times7 + 2\times9$.

Step4: Calculate the result

$\frac{dA}{dt}=56+18=74$ in²/s.

Answer:

74