Arectangle initially has dimensions 6 cm by 7 cm. all sides begin increasing in length at a rate of 1 cm/s. at what rate is the area of the rectangle increasing after 18 s? let a, b, and h be the area, base, and height of a rectangle, respectively. write an equation relating a, b, and h. differentiate both sides of the equation with respect to t. dt dt after 18 s, the area of the rectangle is increasing at a rate of

  49 cm²/s

Step-by-step explanation:

The problem statement tells you what to do.

Write an equation relating A, b, h:

  A = bh . . . . . . the equation for the area of a rectangle

Differentiate with respect to t:

  dA/dt = (db/dt)h + b(dh/dt) . . . . . . . product rule

To find the rate of change after 18 seconds, you need to know the dimensions b and h after 18 seconds. Since each dimension was increasing at the rate of 1 cm/s, it is 18 cm more than it was at the beginning:

At 18 seconds,

  b = 6 cm + 18 cm = 24 cm;

  h = 7 cm + 18 cm = 25 cm.

Of course, db/dt = dh/dt = 1 cm/s. Then the rate of change of area is ...

  dA/dt = (1 cm/s)(25 cm) + (24 cm)(1 cm/s)

  dA/dt = 49 cm²/s

You could write a formula for the area as a function of time and differentiate that:

  A = (6 +t)(7 +t) = 42 + 13t + t²

Then the derivative is ...

  dA/dt = 13 +2t

and when t=18, this is ...

  dA/dt = 13 + 2(18) = 49 . . . . cm²/s

0.84 dang lotta people are looking for this lol or if you round up its 1 but the overall answer to that is 0.84

step-by-step explanation: