A machinery pedestal is made of two concrete cubes (not neccessarily of the same size) one on top of the other. The pedestal is 8.00 ft high and contains 152ft^3 of concrete. Find the edge length, x and y, of each cube. If necessary, round your answers for x and y to two decimal places.

Let x be the side length of one cube, and y the side length of the other.

Since the height of the pedestal is 8ft, we have x+y=8.

Since the volume of the pedestal is 152 cubic feet we have `x^3+y^3=152` (since `x^3,y^3` are the respective volumes of the two...

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Let x be the side length of one cube, and y the side length of the other.

Since the height of the pedestal is 8ft, we have x+y=8.

Since the volume of the pedestal is 152 cubic feet we have `x^3+y^3=152` (since `x^3,y^3` are the respective volumes of the two cubes.)

`x^3+y^3=152` Factor the sum of cubes

`(x+y)(x^2-xy+y^2)=152`  Substitute for x+y:

`8(x^2-xy+y^2)=152`

`x^2-xy+y^2=19`    Add and subtract 3xy to the left side:

`x^2-xy+3xy+y^2-3xy=19`

`x^2+2xy+y^2-3xy=19`  Factor the perfect square trinomial

`(x+y)^2-3xy=19` Substitute for x+y:

`64-3xy=19`

-3xy=-45

xy=15

Now we have x+y=8 and xy=15:

`xy=15 ==> x=15/y`

`15/y+y=8`

`15+y^2=8y`

`y^2-8y+15=0`

`(y-3)(y-5)=0`

y=3 or y=5. If y=3 then x=5, and if y=5 then x=3.

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The side lengths of the cubes are 3ft and 5ft

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