The altitude of a triangle is increasing at a rate of 1.500 centimeters/minute while the area of the triangle is increasing at a rate of 2.000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 10.000 centimeters and the area is 98.000 square centimeters?
We are given that the altitude of the triangle is increasing at a rate of 1.500 centimeters/minute and the area of the triangle is increasing at a rate of 2.000 square centimeters/minute.
Let $A$ be the area of the triangle and $h$ be the altitude of the triangle. Let $B$ be the base of the triangle.
We are asked to find $\frac{dB}{dt}$ when $h = 10.000$ centimeters and $A = 98.000$ square centimeters.
We know that the area of a triangle is given by the formula $A = \frac{1}{2} \times B \times h$. Taking the derivative of both sides with respect to time, we obtain:
$$\frac{dA}{dt} = \frac{1}{2} \times \left(\frac{dB}{dt} \times h + B \times \frac{dh}{dt}\right)$$
Substituting the given values, we have:
$$2 = \frac{1}{2} \times \left(\frac{dB}{dt} \times 10.000 + B \times 1.500\right)$$
Simplifying the equation, we get:
$$4 = \frac{dB}{dt} \times 10.000 + B \times 1.500$$
Rearranging the equation, we find:
$$\frac{dB}{dt} \times 10.000 = 4 - B \times 1.500$$
Substituting $B = \frac{2A}{h}$, we have:
$$\frac{dB}{dt} \times 10.000 = 4 - \frac{2A}{h} \times 1.500$$
Substituting $h = 10.000$ and $A = 98.000$, we get:
$$\frac{dB}{dt} \times 10.000 = 4 - \frac{2 \times 98.000}{10.000} \times 1.500$$
Simplifying the equation, we find:
$$\frac{dB}{dt} \times 10.000 = 4 - 2 \times 9.800 \times 1.500$$
$$\frac{dB}{dt} \times 10.000 = 4 - 29.400$$
$$\frac{dB}{dt} \times 10.000 = -25.400$$
Dividing both sides by 10.000, we obtain:
$$\frac{dB}{dt} = -2.540$$
Therefore, the base of the triangle is changing at a rate of $-2.540$ centimeters/minute when the altitude is $10.000$ centimeters and the area is $98.000$ square centimeters.