Question

What are the dimensions of V=2y^3+17y+8y

Answer 1

   
We decompose expression:


$\displaystyle\\ V=2y^3+17y^2+8y = y(2y^2+17y+8)\\\\ y_{12}= \frac{-17 \pm \sqrt{17^2-4\times2\times8} }{4}= \frac{-17 \pm \sqrt{289-64} }{4}= \\ \\ = \frac{-17 \pm \sqrt{225} }{4}=\frac{-17 \pm 15}{4}\\\\ y_1 = \frac{-17 - 15}{4}=\frac{-32}{4}=\boxed{-8}\\\\ y_2=\frac{-17 +15}{4}=\frac{-2}{4}=\boxed{-\frac{1}{2}}\\\\ \Longrightarrow~~V=y(2y^2+17y+8) = \boxed{y(y+8)(y+\frac{1}{2})}\\\\ \Longrightarrow~~ L = y+8,~~~l = y+\frac{1}{2} ~\text{ and }~ h = y$