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Fernando has two equally-sized containers of clay which he is using to build a pyramid-like structure. The finished structure will consist of four square-based, two-inch thick slabs of clay, with each slab two inches less wide than the one beneath it. Fernando has one-half of a container of clay left after forming the first two slabs. Does he have enough clay to finish the structure?

User Bincy Baby
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Answer:

Fernando does not have enough clay to finish the structure since, the expressions for the volume left are not the same.

Explanation:

Let w be the length of the longest slab. Since each slab is 2 inches less than the one beneath it, we have the width of the other 3 slabs as w - 2, w - 2 - 2 = w - 4 and w - 4 - 2 = w - 6 respectively.

Since the base of each slab is a square and has thickness 2 inches, the volume of each slab from largest to smallest is thus 2w², 2(w - 2)², 2(w - 4)² and 2(w - 6)² respectively.

We have that we have half of the container of clay left after forming the first two slabs(which are the bottom-most slabs). Let V be the volume of each clay container. Since we have half left, that is V/2, we have used 2V - V/2 = 3V/2 to make the two bottom-most slabs.

So, 3V/2 = 2w² + 2(w - 2)²

= 2w² + 2(w² - 4w + 4)

= 2w² + 2w² - 8w + 8

= 4w² - 8w + 8

= 4(w² - 2w + 2) (1)

Also we want to know if V/2 is equal to the volume of the two top-most slabs.

So, V/2 = 2(w - 4)² + 2(w - 6)²

= 2(w² - 8w + 16) + 2(w² - 12w + 36)

= 2w² - 16w + 32 + 2w² - 24w + 72

= 4w² - 40w + 104

= 4(w² - 10w + 26)

From (1) V/2 = 4(w² - 2w + 2)/3 ≠ 4(w² - 10w + 26)

Since the expressions for the remaining volume are not the same, Fernando does not have enough clay to finish the structure.

User Kieranties
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