The altitude of a triangle is increasing at a rate of 3.000 centimeters/minute while the area of the triangle is increasing at a rate of 4.500 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 12.000 centimeters and the area is 85.000 square centimeters?
Note: The "altitude" is the "height" of the triangle in the formula "Area=(1/2)*base*height". Draw yourself a general "representative" triangle and label the base one variable and the altitude (height) another variable. Note that to solve this problem you don't need to know how big nor what shap

Question

The altitude of a triangle is increasing at a rate of 3.000 centimeters/minute while the area of the triangle is increasing at a rate of 4.500 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 12.000 centimeters and the area is 85.000 square centimeters?
Note: The "altitude" is the "height" of the triangle in the formula "Area=(1/2)*base*height". Draw yourself a general "representative" triangle and label the base one variable and the altitude (height) another variable. Note that to solve this problem you don't need to know how big nor what shap

dumbcow · Answer

$$\frac{dA}{dt} = \frac{dh}{dt}*\frac{dA}{dh}+\frac{db}{dt}*\frac{dA}{db}$$
solve for db/dt
$$\frac{db}{dt} = \frac{9-3b}{h}$$