Related Rates: The altitude of a triangle is increasing at a

Math-hating-Aggie

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The altitude of a triangle is increasing at a rate of 4 cm/min, while its area is increasing at a rate of 1 sq. cm/min. At what speed is the base of the triangle changing when the altitude of the triangle is 3 cm and its area is 18 sq. cms?

This was a multiple choice question and my answer was 1/2 but it wasn't in the given options
 
You probably know the area of a triangle is \(\displaystyle A=\frac{1}{2}bh\)

Let b=base and h=height(altitude).

Differentiate:

\(\displaystyle \L\\\frac{dA}{dt}=\frac{1}{2}[b\frac{dh}{dt}+h\frac{db}{dt}]\)

When the height is 3 and the area is 18, then the base is 12.

Now you have b, h, dh/dt, dA/dt. Solve for db/dt.
 
Please Note: The area is increasing more slowly (1 of its units) than the altitude (4 of its units). Since this is a direct, joint variation problem, the base had better be shrinking. In other words, db/dt < 0!! If you get db/dt >0, go back and find your error.
 
Math-hating-Aggie said:
...my answer was 1/2 but....
It is difficult to find errors we cannot see. Please reply showing how you arrived at your answer.

Thank you.

Eliz.
 
Additional Note: If you are struggling with mathematics, please review your screen name for a hint on how to improve your performance. Please consider getting a better attitude.
 
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