Thread: Dimensions

1. Dimensions

A rectangular in-ground pool is to be installed. The engineer overseeing the construction project estimates that at least 1408m^3 of earth and rocks needs to be excavated. What are the minimum dimensions of the excavation if the depth must be 2m more than one quarter the width and the length must be 12m more than four times the width?

is l x w x d : 44 m by 8 m by 4 m the correct dimensions?

Please help me check. Thanks!!

2. Hello, Hellooo!

A rectangular in-ground pool is to be installed.
The engineer estimates that at least 1408 m³ need to be excavated.
What are the minimum dimensions of the excavation
if the depth must be 2m more than one quarter the width
and the length must be 12m more than four times the width?

$\displaystyle \text{Is this correct? }\;L \times W \times D \:=\: 44\:m \times 8\:m \times 4\:m$ . Yes!
This is how I solved it . . .

We are told that: .$\displaystyle \begin{Bmatrix}L &=& 4W + 12 \\ W &=& W \\ D &=& \frac{1}{4}W + 2 \end{Bmatrix}$

Since $\displaystyle V \:=\:\L\cdot W\cdot D \:=\:1408$

. . we have: .$\displaystyle (4W + 12)\cdot W \cdot(\frac{1}{4}W + 2) \;=\;1408$

which simplifies to: .$\displaystyle W^3 + 11W^2 + 24W - 14087 \;=\;0$

. . which factors: .$\displaystyle (W - 8)(W^2 + 19W + 176) \;=\;0$

and has one real root: .$\displaystyle W \:=\:8$

Therefore: .$\displaystyle \begin{Bmatrix} L &=& 44 \\ W &=& 8 \\ D &=& 4 \end{Bmatrix}$