If a rectangle has a base of 2cm, and length of 3cm, and its base increases by 1cm/s and its length increases by 2cm/s, at what rate is the area increasing when the area is 28cm^2?

Is the answer to this 15cm^2/s?

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- March 4th 2010, 02:30 PMBrownianManArea of rectangle
*If a rectangle has a base of 2cm, and length of 3cm, and its base increases by 1cm/s and its length increases by 2cm/s, at what rate is the area increasing when the area is 28cm^2?*

Is the answer to this 15cm^2/s? - March 4th 2010, 04:03 PMBladeowen
Ok

Think of the graph of time on x axis and length on y axis plot the lines of length and base:

(y=mx+c where c is value when t=0 and m is the rate of change of length)

Length y = 2+x

Base y= 3+2x

So

Area = (2+x)(3+2x)

solve 28=(2+x)(3+2x) to give 2 values for time when area=28.

One of these values are negative, and this clearly is invalid as you can't have negative time.

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Now the area at any point in time is (2+x)(3+2x) as shown above.

Put this equal to y and find "dy by dx"

Put your realistic value for time in here to get the rate of change of area at this point :)

It comes out at 15cm^2/s

**EDIT:Aha i didn't read your post properly - you just wanted confirmation - well yea its 15 - sorry :)** - March 4th 2010, 04:27 PMBrownianMan
Thanks!

This was a question on a test I just had. Glad to get confirmation of its accuracy! :)