# Area Increasing

• Jan 23rd 2014, 01:05 PM
JenniferC
Area Increasing
The length of a rectangle is increasing at a rate of 2 ft. per second, while the width is increasing at a rate of 3 ft. per second. When the length is 4 ft. and the width is 6 ft., how fast, in square feet per second, is the area increasing?

I solved:

A = LW
dA/dt = LdW/dt + WdL/dt
dA/dt = L*1 + W*2
at L = 4, W = 6, dA/dt = 4+6*2 = 16 ft²/sec

...but that's not an answer choice. The answer choices are 144, 24, 5, 26, and 6.

I'd really appreciate some some on this, possibly clear steps on how to solve.
• Jan 23rd 2014, 01:21 PM
romsek
Re: Area Increasing
Quote:

Originally Posted by JenniferC
The length of a rectangle is increasing at a rate of 2 ft. per second, while the width is increasing at a rate of 3 ft. per second. When the length is 4 ft. and the width is 6 ft., how fast, in square feet per second, is the area increasing?

I solved:

A = LW
dA/dt = LdW/dt + WdL/dt
dA/dt = L*1 + W*2
at L = 4, W = 6, dA/dt = 4+6*2 = 16 ft²/sec

...but that's not an answer choice. The answer choices are 144, 24, 5, 26, and 6.

I'd really appreciate some some on this, possibly clear steps on how to solve.

I think you meant

$\frac{dA}{dt}=L\cdot 3+W\cdot 2$

at L=4 and W=6

$\frac{dA}{dt}=4\cdot 3+6\cdot 2=12+12=24$