Math (:
posted by <3 .
The altitude of a triangle is increasing at a rate of 2.500 centimeters/minute while the area of the triangle is increasing at a rate of 1.500 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 10.500 centimeters and the area is 93.000 square centimeters?

Given
A=bh/2, b=2*93/10.5=17.714
differentiate with respect to t (by the product rule on the right hand side).
The only unknown left is the rate of change of the base.
Hint: it is negative. 
let the area be A
let the base be x
let the height be y
given: at a time of t minutes,
dA/dt = 1.5 cm^2/min
dy/dt = 2.5 cm^2/min
find:
dx/dt when A = 93 and y = 10.5
A = xy/2 or
2A = xy (equ#1)
differentiate implicitly with respect to t, using the product rule
2dA/dt =x(dy/dt) + y(dx/dt) (equ#2)
we know A=93 when y = 10.5, so x = 17.714
sub into equ#2
2(1.5) = 17.714(2.5) + 10.5dx/dt
solve for dx/dt 
My answer is 3.9319, but it's wrong.

I have 3.932 too.
Can you check the numbers in the question? 
I copied & pasted this from my online homework. It said the answer is wrong.

I also had 3.932
What answer did the book have? 
I use an online homework program called WebWork. It just tells you if your answer is correct or not.

Try 3.93, 3.932.

The answer is 3.9300. Thanks!