# The altitude of a triangle is increasing at a rate of 1.5 cm/min while the area of the triangle is increasing at a rate of 5 square cm/min. At what rate is the base of the triangle changing when the altitude is 9 cm and the area is 81 square cm?

This is a related rates (of change) type problem.

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The altitude of a triangle is increasing at a rate of 1.5 cm/min while the area of the triangle is increasing at a rate of 5 square cm/min. At what rate is the base of the triangle changing when the altitude is 9 cm and the area is 81 square cm?
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The variables of interest are
##a## = altitude
##A## = area and, since the area of a triangle is ##A=1/2ba##, we need
##b## = base.

The given rates of change are in units per minute, so the (invisible) independent variable is ##t## = time in minutes.

We are given:
##(da)/dt = 3/2## cm/min

##(dA)/dt = 5## cm##””^2##/min

And we are asked to find ##(db)/dt## when ##a = 9## cm and ##A = 81##cm##””^2##

##A=1/2ba##, differentiating with respect to ##t##, we get:

##d/dt(A)=d/dt(1/2ba)##.

We’ll need the on the right.

##(dA)/dt = 1/2 (db)/dt a + 1/2b (da)/dt##

We were given every value except ##(db)/dt## (which we are trying to find) and ##b##. Using the formula for area and the given values of ##a## and ##A##, we can see that ##b=18##cm.

Substituting:

##5= 1/2 (db)/dt (9)+1/2(18)3/2##

Solve for ##(db)/dt = -17/9##cm/min.

The base is decreasing at ##17/9## cm/min.

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