Posted by **Anonymous** on Tuesday, July 12, 2011 at 5:00pm.

The altitude of a triangle is increasing at a rate of 2500 centimeters/minute while the area of the triangle is increasing at a rate of 4000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 11500 centimeters and the area is 84000 square centimeters?

- Calc -
**Reiny**, Tuesday, July 12, 2011 at 5:26pm
Area = (1/2) bh , where b is the base and h is the height

d(Area)/dt = (1/2)b dh/dt + (1/2)h db/dt

when h = 11500 , A = 84000

84000 = (1/2)(b)(11500)

b = 14.609

in derivative equation ....

4000 = (1/2)(14.609)(2500) + (1/2)(11500)db/dt

db/dt = (8000 - 14.609(2500))/11500

= -2.48 cm/min

check my arithmetic, the negative seems to suggest that at that moment the base is decreasing.

- Calc -
**Max**, Tuesday, July 12, 2011 at 5:27pm
Try considering the formula for the area of a triangle: A = (1/2)bh.

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