1. ## Area of Rectangle

The initial dimensions of a rectangle are 5x7 cm and the length and width of the reactangle are increasing by the rate of 2 cm/sec. How long will it take for the area to be at least 5 times its initial size?

I use a calculator to figure out when the size will be equal to or greater to 175 cm squared and the time would be atleast 3 minutes. But how do i figure out the time to the nearest second, the exact time in seconds?

2. ## this help?

175<(5+2t)(7)

Where t is time

3. Originally Posted by chrozer
The initial dimensions of a rectangle are 5x7 cm and the length and width of the reactangle are increasing by the rate of 2 cm/sec. How long will it take for the area to be at least 5 times its initial size?

I use a calculator to figure out when the size will be equal to or greater to 175 cm squared and the time would be atleast 3 minutes. But how do i figure out the time to the nearest second, the exact time in seconds?
InitialArea = 5 x 7 = 35

Each side increases at the rate of 2cm/sec
Thus,

The Area at time zero = (5+2*0)(7+2*0) = 35
the area at time of 1 second: (5+2*1)(7+2*1) = 7*9 = 63
the area at time of 2 seconds: (5+2*2)(7+2*2) = 9*11 = 99
the area at time s will be: (5+2*s)(7+2*s)

What we seek is a rectangle that is 5 times the initial area.
5*35 = 175

the area at time in seconds, s : (5+2*s)(7+2*s) = 175
expanding:
5*7 + 10*s + 14*s + 4*s^2 = 175
simplifying:
4s^2 + 24s - 140 = 0

A simple quadratic equation, solve for s.
The actual answer is between the 3rd second and the 4th second.

4. Originally Posted by aidan
InitialArea = 5 x 7 = 35

Each side increases at the rate of 2cm/sec
Thus,

The Area at time zero = (5+2*0)(7+2*0) = 35
the area at time of 1 second: (5+2*1)(7+2*1) = 7*9 = 63
the area at time of 2 seconds: (5+2*2)(7+2*2) = 9*11 = 99
the area at time s will be: (5+2*s)(7+2*s)

What we seek is a rectangle that is 5 times the initial area.
5*35 = 175

the area at time in seconds, s : (5+2*s)(7+2*s) = 175
expanding:
5*7 + 10*s + 14*s + 4*s^2 = 175
simplifying:
4s^2 + 24s - 140 = 0

A simple quadratic equation, solve for s.
The actual answer is between the 3rd second and the 4th second.
Ok I see... now. Yeah I meant between the 3rd and 4th second not minute.

Ok I solved it on the calculator and I got around 3.633 seconds. Thanks alot.