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?

Question

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?

Reiny · Answer

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.

Max · Answer

Try considering the formula for the area of a triangle: A = (1/2)bh.