SOLUTION: A cubic shipping container had a volume of v3 cubic meters. The height of the container was decreased by a whole number of meters and the width was increased by a whole number of m
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Question 251549: A cubic shipping container had a volume of v3 cubic meters. The height of the container was decreased by a whole number of meters and the width was increased by a whole number of meters so that the volume of the container is now v3 + 2v2 – 3v. Find out the following:
a. By how many meters the height was decreased?
b. By how many meters the width was increased?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
v[1] = v^3
v[2] = v^3 + 2v^2 - 3v
let a = amount height was decreased.
let b = amount width was increased.
v * (v-a) * (v+b) = v^3 + 2v^2 - 3v.
find the roots of v^3 + 2v^2 - 3v.
set the equation equal to 0 to get:
v^3 + 2v^2 - 3v = 0
factor out a v to get:
v * (v^2 + 2v - 3) = 0
factor v^2 + 2v - 3 to get:
(v-1) * (v+3) * v = 0
you get:
a = 1
b = 3
confirm by substituting in original equation to get:
v * (v-1) * (v+3) = (v^2 -v) * (v+3) = v^3 + 3v^2 -v^2 - 3v.
combine like terms to get:
v * (v-1) * (v+3) = v^3 + 2v^2 - 3v
since this is identical to the original equation of v^3 + 2v^2 - 3v, then the values for a and b are good.
v*v*v = v^3 = original equation for the cube.
v*(v-1)*(v+3) = v^3 + 2v^2 - 3v = original equation after taking 1 unit off the length and adding 3 units to the width.
answer is:
height was decreased by 1 unit.
width was increased by 3 units.
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