Related Rates - A Rectangle is Expanding

Question

A 9 by 4 centimeter rectangle expands in such a way that the 9–centimeter side is expanding at the rate of 5 centimeters per second and the proportions of the rectangle remain constant. At what rate is the area of the rectangle increasing when its dimensions are 36 by 16 centimeters?



A. 120 cm²/s 

B. 140 cm²/s 

C. 160 cm²/s 

D. 180 cm²/s

Answer 1

The correct answer is C = 160

First, you need to determine at which speed the second side is increasing so that the PROPORTIONS of the rectangle stay the same

So one side is 9+5t (t is time in seconds)

The other side is 4+ R*t (the unknown Rate)

The ratio is 9/4:

9+5t = 9/4(4+Rt)

Solving: R=20/9

Now, the Area as a function of t is

A(t) = (9+5t)(4+20/9*t) = 36 + 40t + 100/9 t^2´

The rate of change is A^´

40 + 100/9 * 2t

We are interested in the specific time, when the size of (9+5t) = 36, so t = 27/5

Pluging it in:

40 + 100/9 * 2 * 27/5 = 160

Answer 2

Let A=xy so dA/dt = xdy/dt + ydx/dt and let x be the 4 cm side and y be the 9 cm side. So dy/dt =5 cm/sec and since 9/4 =y/x then 9/4 =dy/dt/ dx/dt and dx/dt = 20/9 cm/sec. Using this in the dA/dt equation gives dA/dt= 16•5 + 36•20/9 = 80 +80 = 160 cm²/sec or C.