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?
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.
To determine the side length of the small base that minimizes the total lateral area of the truncated square pyramid, we need to find an equation that describes the lateral area in terms of the side length of the small base.
The lateral area of a truncated square pyramid consists of the areas of the four trapezoidal faces. The formula for the area of a trapezoid is:
Area = (1/2) * (b1 + b2) * h
In this case, the height of the trapezoidal faces is the same as the height of the truncated square pyramid, which is 30 cm.
Let's denote the side length of the small base as x cm. The side length of the large base is given as 10 cm.
To find the lengths of the bases of the trapezoidal faces, we can use similar triangles. By drawing a perpendicular from the vertex of the small base to the large base, we create two similar triangles.
The ratio of the corresponding sides of similar triangles is equal. Therefore, we can write:
x / (10 - x) = 30 / 10
Now, we can solve this equation to find the value of x.
Cross-multiplying the equation gives us:
x * 10 = 30 * (10 - x)
10x = 300 - 30x
40x = 300
x = 7.5 cm
Therefore, the side length of the small base that minimizes the total lateral area of the truncated square pyramid is 7.5 cm.