Math calculus
posted by Kitor .
The altitude of a triangle is increasing at a rate of 1000 centimeters/minute while the area of the triangle is increasing at a rate of 1500 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 7000 centimeters and the area is 89000 square centimeters?
I have tried to answer this in so many ways but couldn't get right answer ... Help please !

Let the base be x cm and the height be y cm
Area = (1/2)xy cm^2
dArea/dt = (1/2)(x dy/dt + y dx/dt)  #1
given: dy/dt = 1000 cm/min
dArea/dt = 1500 cm^2/min
find dx/dt when y = 7000 cm and Area = 89000 cm^2
...........
when y = 7000 and area = 89000
(1/2)(x)(7000) = 89000
x = 178/7
which leaves dx/dt as the only missing part of #1
1500 = (1/2)( (178/7)(1000) + 7000(dx/dt) )
3000 = 178000/7 + 7000 dx/dt
dx/dt = 157000/49000 cm/min
=  157/49 cm/min
= appr 3.2 cm/min
At that instant, the base is decreasing at a rate of appr 3.2 cm/min