Richard cuts a $ 4\frac{1}{2} ~\text{ft} \times 7\frac{1}{2} ~\text{ft} \times 11\frac{1}{4}~ \text{ft}$ rectangular prism into congruent cubes without any part of the prism leftover. If he makes the cubes as large as possible, how many will he produce?
+14913 How many will he produce?
Hello Guest!
Richard cuts a \( 4\frac{1}{2} ~\text{ft} \times 7\frac{1}{2} ~\text{ft} \times 11\frac{1}{4}~ \text{ft}\) rectangular prism into congruent cubes without any part of the prism leftover. If he makes the cubes as large as possible, how many will he produce?
Richard divides the length, width and height of the cube into two halves each. In this way he produces with just three cuts 2 times 2 times 2 equals 8 cubes.
!
+128460 Let us make the edges of the cube 100 times as large so that we have a prism of
450, 750 and 1125 (this prism will be similar to the smaller one )
The greatest commom divisor of the three dimensions = 75 = the edge length of each cube
On the top of the prism we can divide one dimension into 450/75 = 6 parts
And, again on the top, we can divide another dimension into 750/75 = 10 parts
So, we have 6 * 10 = 60 cubes in each layer
And we can divide the remaining dimension into 1125/75 = 15 layers (parts)
So....the max number of cubes = 6 * 10 * 15 = 900
