The problem you're working on asks to find the dimensions of a triangle with a base and height that add up to 16 cm, where the area is maximized.

Let's break this down:

Problem:

• The sum of the base and height of a triangle is 16 cm.
• The area of a triangle is given by: $A = \frac{1}{2} \times \text{base} \times \text{height}$
• We are asked to find the dimensions (base and height) that maximize the area.

Step-by-step solution:

1. Let the base of the triangle be $b$ and the height be $h$. According to the problem, $b + h = 16$. So, $h = 16 - b$.

2. Substitute $h = 16 - b$ into the area formula: $A = \frac{1}{2} \times b \times (16 - b)$ $A = \frac{1}{2} \times (16b - b^2)$ $A = 8b - \frac{1}{2}b^2$

3. To find the maximum area, we take the derivative of $A$ with respect to $b$, and set it equal to zero: $\frac{dA}{db} = 8 - b = 0$ Solving this gives $b = 8$.

4. Since $b + h = 16$, if $b = 8$, then $h = 16 - 8 = 8$.

Conclusion:

The triangle with maximum area has a base of 8 cm and a height of 8 cm.

Would you like any additional details or further explanation?

Here are 5 related questions:

1. How do you find the maximum of a function using calculus?
2. Why does the triangle with equal base and height maximize the area?
3. What are other methods to find maximum or minimum values in geometry problems?
4. How do constraints affect optimization problems in general?
5. What is the significance of the derivative in finding maxima and minima?

Tip: In optimization problems, always start by expressing one variable in terms of the other using the given constraints!