A trough, shown in the figure, is 64 ft long and its ends have the shape of isosceles triangles whose sides are 2 ft long.

Find the length of the base of the two isosceles triangles, marked \(x\) in the figure, that maximizes the volume.

Let V be the volume of the trough. Then

V=(Area of the isosceles triangle)\(\cdot 64\)

Let A be the area of the isosceles triangle.

Then \(A=\frac {1}{2}\cdot x\cdot h\),

and

\(A=\frac {1}{2}\cdot x\cdot \sqrt {4-\frac {x^2}{4}}\)

\[ x = \answer {2\sqrt {2}} \]