The problem involves finding the dimensions of a rectangle with the largest possible area inscribed in a triangle. Here's the breakdown and solution:

Problem Setup:

1. A triangle with base $b$ and height $h$ is given.
2. A rectangle is inscribed in the triangle such that:
   • Two of the rectangle's vertices lie on the base of the triangle.
   • The other two vertices lie on the two sides of the triangle.
3. We need to find the dimensions of the rectangle that maximizes its area.

Solution:

1. Relationship Between Variables:

   • Place the triangle in the coordinate system with its base along the $x$-axis and the vertex opposite the base at $(b/2, h)$.
   • Assume the rectangle has a width $w$ and height $y$ where the upper vertices of the rectangle lie on the sides of the triangle.

2. Equations for the Triangle's Sides:

   • The equations of the sides of the triangle are:
     • Left side: $y = \frac{2h}{b} x$
     • Right side: $y = -\frac{2h}{b} x + 2h$

3. Maximizing the Rectangle Area:

   • Let the base of the rectangle be $2x$, with the rectangle's vertices at $(x, y)$ and $(-x, y)$.
   • The rectangle's area is $A = 2x \cdot y$.
   • Substitute $y$ from the line equation $y = \frac{2h}{b} x$: $A = 2x \cdot \frac{2h}{b} x = \frac{4h}{b} x^2$

4. Optimization:

   • To maximize $A$, take the derivative with respect to $x$: $\frac{dA}{dx} = \frac{8h}{b} x$
   • Set $\frac{dA}{dx} = 0$ to find critical points: $x = 0 \quad \text{or maximum at } x = b/4$

5. Dimensions of the Maximum Rectangle:

   • If $x = b/4$, substitute back into the equation for $y$: $y = \frac{2h}{b} \cdot \frac{b}{4} = \frac{h}{2}$
   • The rectangle's dimensions are:
     • Width: $2x = \frac{b}{2}$
     • Height: $y = \frac{h}{2}$

6. Maximum Area: $A_{\text{max}} = \frac{4h}{b} \left( \frac{b}{4} \right)^2 = \frac{bh}{4}$

Final Answer:

The dimensions of the rectangle with the largest area are:

• Width: $\frac{b}{2}$
• Height: $\frac{h}{2}$

Would you like further clarification or details on any specific part of the process? Here are five related questions to consider:

1. How is the optimization process carried out using derivatives?
2. What other methods can be used to find maximum areas in similar geometric problems?
3. Could a different placement of the rectangle lead to a larger area?
4. How would the solution change if the triangle were scalene?
5. Can the same principles be applied to 3D geometry, such as maximizing the volume of a rectangular prism inside a pyramid?

Tip: When optimizing areas or volumes in geometry, always check the constraints carefully, as they dictate the limits of your solution.