The Minimum Gift Wrapper Problem

The last problem in the previous post asked about the least wrapper that can be used to wrap the box below. As we can see, if we wrap each side with a rectangular wrapper, it is equivalent to finding the areas of all the faces or finding the surface area of the prism. The original problem is as follows.

The Minimum Gift Wrapper Problem

Rina placed a gift for her friend in a box whose dimensions are shown below. She wants to cover the box. What is the least area of the wrapper that she could use? 


If we look at the box, there are 3 pairs of faces (this means 6 faces in all). The front face (identical with the back face), the left face (identical with the right face), and the top face (identical with the bottom face). So, we just get the areas of the front, right, and top faces, and multiply each by 2.

Area of front and back faces = 6 × 12 × 2 = 144 

Area of the  top and bottom faces =  4 × 12 × 2 = 96

Area of the right and left faces =  6 × 4 × 2 = 48

So, the least wrapper area that can be used is 144 + 96 + 48  = 288 sq. cm

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