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.
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?
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