Explanation::

Since the rectangular box is 12 * 18 * 10

hence the soup cans which is of diameter 6 and height 5 can be arranged

as : 3 rows of 2 each i.e 6 cans

but the height of the can is 5 and the total height of the rectangular box is 10,

so an additional slot of 6 cans will fit inside

This led to a total of = 12 cans
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Now, the cans can be placed along either of 3 sides.
But we should choose a side that will not leave any gap or minimum gaps.
Out of three sides 12, 18 and 10, the side 10 is a multiple of 5, so two rows/columns will fit in exactly.
Even the base in that scenario 12*18 will be exactly covered by the dia 6 of the can.

so along side of 12, there can be 12/6 or 2 columns, and along the side 18, we will have 18/6 or 3 columns. Finally along height 10, we will have 10/2 or 2 rows.
Thus, total 2*3*2=12
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The question comes down to determining how to orient the box. Do, we make the HEIGHT of the box 12 inches, 18 inches or 10 inches?
Well, since the height of each CAN is 5 inches, we can see that making the HEIGHT of the box 10 inches is a great course of action. Otherwise, there will be empty space above the cans.

Also, if the radius of each can is 3 inches, then the DIAMETER = 6 inches
So, we want the base of the box to be such that we can fit as many 6-inch diameters into it.
Since 12 inches and 18 inches are both multiples of 6 inches, we can maximize the number of cans by making the 12-inch by 18-inch part of the box the base.

Since with a 12-inch by 18-inch base, we can place 3 rows of 2 cans on the base. This is a total of 6 cans (so far)
Then, we can place a second level of 6 cans on top of the first 6 cans.

So, the most cans we can place in the box = 6 + 6 = 12

Answer: 12

Cheers,
Brent
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Notice that each cylinder will take up an area of 6 x 6 = 36 square inches on the base of the box, even though the base area of each cylinder is less than that. Therefore, we can think of the soup cans as 6 x 6 x 5 rectangular boxes instead of cylinders. If the big box is laid on the 12 x 18 base, there will be 12/6 = 2 rows of cans and 18/6 = 3 columns of cans; totaling 2 x 3 = 6 cans on the base. Since the height is 10, 10/5 = 2 cans be stacked; therefore 6 x 2 = 12 cans is the maximum number to fit in the box.
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Here is an illustrated explanation.

Fantastic illustration! Really needed the visualization

I concur!!
Great work, arc601!!
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I like this optimization problem.

fitting cans with diameter of 6 in. and height of 5 in. should not be difficult, if we can choose constraints for box width by length = 12 by 18 and height = 10

width and length wise fitting is \(12/6*18/6=2*3=6\) and height wise fitting is \(10/5=2\)

hence \(6*2\) cans can fit into the box

answer is 12

Also, IMO the other way to put this problem is that on the surface of a box we need to have squares with side=6 to fit the diameter=6. Recall that width by length is equal to \(12*18=216\) and \(216=6^3\)

What is \(6^3\) ?

\(6^3\) implies that we have \(6*6^2\), which is alternatively 6 squares each with side=6. Cool?