We are given a triangle where the base is increasing and the altitude is decreasing, and we need to find the rate at which the area of the triangle is changing at a specific moment.

Step 1: Formula for the Area of a Triangle

The area $A$ of a triangle is given by: $A = \frac{1}{2} \times \text{base} \times \text{height}$ Let:

• $b$ = base of the triangle (in inches),
• $h$ = height or altitude to the base (in inches),
• $A$ = area of the triangle (in square inches).

Thus, the area $A$ at any time $t$ is: $A(t) = \frac{1}{2} \times b(t) \times h(t)$

Step 2: Differentiate the Area with Respect to Time

To find how fast the area is changing, we take the derivative of the area with respect to time $t$: $\frac{dA}{dt} = \frac{1}{2} \left( \frac{db}{dt} \cdot h + b \cdot \frac{dh}{dt} \right)$ Where:

• $\frac{db}{dt}$ is the rate of change of the base,
• $\frac{dh}{dt}$ is the rate of change of the height.

Step 3: Plug in the Given Values

From the problem, we are given:

• $\frac{db}{dt} = 13$ inches per second (the base is increasing),
• $\frac{dh}{dt} = -14$ inches per second (the height is decreasing),
• $b = 18$ inches,
• $h = 6$ inches.

Now, substitute these values into the differentiated equation: $\frac{dA}{dt} = \frac{1}{2} \left( 13 \cdot 6 + 18 \cdot (-14) \right)$ Simplifying: $\frac{dA}{dt} = \frac{1}{2} \left( 78 - 252 \right)$ $\frac{dA}{dt} = \frac{1}{2} \times (-174)$ $\frac{dA}{dt} = -87 \, \text{square inches per second}$

Step 4: Conclusion

The area of the triangle is decreasing at a rate of $87 \, \text{square inches per second}$ when the base is 18 inches and the height is 6 inches.

Would you like further clarification or details on any of the steps?

Here are 5 related questions to expand on this concept:

1. How would the rate of change of the area be affected if the altitude were also increasing?
2. What happens to the rate of area change when the base and altitude change at the same rate?
3. How would you find the rate of change of the perimeter of the triangle?
4. Can the rate of change of the area ever become positive in this scenario?
5. How does the chain rule apply in solving related rates problems like this?

Tip: In related rates problems, always differentiate the relevant equation with respect to time before plugging in values.