if the small base has side 2x, then each face of the pyramid is a trapezoid with area
a = (2x+10)/2 * √((5-x)^2 + 30^2)
So, to find the minimum area, da/dx=0
Unfortunately, this never happens. Have I set it up wrong?
Since the height of each face is constant, and the bottom base is constant, naturally its area is least when it is just a triangle: top base=0.
A truncated square pyramid (also known as a frustum) is an object obtained by cutting a square pyramid by a plane parallel to its base (as shown in the figure). For a truncated square pyramid it is known that the side length of the large base is 10 cm and the height (distance between two bases) is 30 cm. For which side length of the small base the total lateral area would be minimized?
1 answer
1 answer