Chapter 4.5, Problem 63E

To determine

To find: the dimensions of given lamp.

Answer to Problem 63E

The dimension of right ramp is $\text{12ft×5ft×}\frac{\text{5}}{\text{3}}\text{ft}$ .

Explanation of Solution

Given information:

The left ramp is twice as long as the right ramp. If 150 cubic feet of concrete are used to build the ramps.

  

Calculation:

Let the length of right ramp is $y$ . So, the length of left ramp is $2y$ .

Setup the equation as,

  $\begin{array}{c}y+3x+2y=21x+6\\ 3y=18x+6\\ y=6x+2\end{array}$

The volume of left ramp is,

  $\begin{array}{c}V=\frac{1}{2}\cdot 2\left(6x+2\right)\left(x\right)\left(3x\right)\\ =3x^2\left(6x+2\right)\\ =18x^3+6x^2\end{array}$

The volume of right ramp is,

  $\begin{array}{c}V=\frac{1}{2}\left(6x+2\right)\left(x\right)\left(3x\right)\\ =\frac{3}{2}{x}^2\left(6x+2\right)\\ =9x^3+3x^2\end{array}$

Because the total volume is 150, so

  $\begin{array}{c}18x^3+6x^2+9x^3+3x^2=150\\ 27x^3+9x^2-150=0\\ 3\left(3x-5\right)\left(3x^2+6x+10\right)=0\\ x=\frac{5}{3}\end{array}$

Because $3x^2+6x+10$ has negative discriminant, so this equation has no real solution.

Therefore, the dimension of right ramp is $\text{12ft×5ft×}\frac{\text{5}}{\text{3}}\text{ft}$ .