A triangle has a base 12 feet long and an altitude 8 feet high. Find the area of the largest rectangle that can be inscribed in the triangle so that the base of the rectangle falls on the base of the triangle.

Make a sketch, that always helps.

Let the length of half the base of the rectangle be x and the height of the rectangle be y.

Let the vertices of the triangle be A, B and C; A and B being at the base. Let the vertices of the rectangle be P, Q, R and S, P and Q being at the base. And then, let X be the midpoint of AB.

From the information given and your sketch, you should see that:

\frac{CX}{AX} = \frac{PS}{AX - PX}

\frac{8}{12} = \frac{y}{6 - x}

\frac{2}{3}(6-x) = y

The area of the rectangle is given by:

A = y(2x) = 2xy

The optimum of A is obtained when the derivative of A is equal to 0.

Taking the two equations, we first get A in terms of y only:

A = 2x(\frac{2}{3}(6-x)) = \frac{4x}{3}(6-x)

Now, find the derivative of A, set that to 0 and solve for x, the length which is half the base of the rectangle. Then you can find the area of the rectangle.

Can you post what you get? :)