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?

Question

Emilia Aguilar · Accepted Answer

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

Charles Anctil · Answer

undefined To find the rate at which the base of the triangle is changing when the altitude is 9 cm and the area is 81 square cm, we can use the formula for the area of a triangle:

Area = (1/2) * base * altitude.

Then, we differentiate both sides of the equation with respect to time t:

d(Area)/dt = (1/2) * (d(base)/dt) * altitude + (1/2) * base * (d(altitude)/dt).

Given that d(Area)/dt = 5 square cm/min, altitude = 9 cm, and d(altitude)/dt = 1.5 cm/min, we can solve for d(base)/dt:

5 = (1/2) * (d(base)/dt) * 9 + (1/2) * base * 1.5.

We also know that when the area is 81 square cm, the altitude is 9 cm. Substituting these values:

81 = (1/2) * base * 9.

From this equation, we can solve for the base, which is 18 cm.

Now, substituting the value of base into the equation for d(base)/dt:

5 = (1/2) * (d(base)/dt) * 9 + (1/2) * 18 * 1.5.

Solving for d(base)/dt:

5 = (4.5/2) * (d(base)/dt) + 13.5.

d(base)/dt = (5 - 13.5) / (4.5/2) = -16/4.5 = -3.56 cm/min.

So, the rate at which the base of the triangle is changing when the altitude is 9 cm and the area is 81 square cm is approximately -3.56 cm/min.