question archive The specifications for a storage container state that the length is 1 metre more than triple the width and the height is 5 metres less than double the width
The specifications for a storage container state that the length is 1 metre more than triple the width and the height is 5 metres less than double the width
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The specifications for a storage container state that the length is 1 metre more than triple the width and the height is 5 metres less than double the width. Find the range of possible dimensions for a volume of at least 8436 m 3 using the algebraic method
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Given the following data
The length is 1 metres more than triple the width
The height is 5 metres less than double the width
Volume of at least 8436 m3
Solution
Using the algebraic method
Let the width be y m
Therefore
The length is 1 metres more than triple the width = 3y + 1 m
The height is 5 metres less than double the width = 2y - 5 m
Volume = L x W x H
Volume of at least 8436 m3
L is the length = 3y + 1 m
W is the width = y m
H is height = 2y - 5 m
(3y + 1 ) x (y ) x (2y - 5) > 8436 m3
y(3y + 1 ) (2y - 5 m) > 8436 m3
y (6y2 - 15y + 2y - 5) > 8436 m3
y (6y2 - 13y - 5) > 8436 m3
6y3 > 13y2 + 5y + 8436
(y - 12)(y2 + 59y/6 + 703/6) > 0
(y - 12) > 0
y > 12
But y is the width
Therefore
Any width is greater than 12 m (Width > 12 m) will give the range of possible dimensions for a volume of at least 8436 m3
