Chapter 4, Problem 28CR

To determine

To Find:The dimensions of the sculpture.

Answer to Problem 28CR

  $6\times 2\times 20$

Explanation of Solution

Given:You use $240$ cubic inches clay to make a sculpture shaped in rectangular prism. The width is four inches less than the length and height is $2$ inches more than three times of the length.

Let the length of the sculpture be $x$ .

Then

The width is four inches less than the length.

Thus

  $width=x-4$

Height is $2$ inches more than three times of the length.

  $height=3x+2$

Now

  $\begin{array}{l}V=240\\ lbh=240\\ x(x-4)(3x+2)=240\\ solving\\ (x^2-4x)(3x+2)-240=0\\ 3x^3-10x^2-8x-240=0\end{array}$

Now using the synthetic division to factor the function we get,

  

Thus

  $\begin{array}{l}(x-6)(3x^2+8x+40)=0\\ \Delta =8^2-4(3)(40)=-416<0\\ x=6\\ now\\ x-4=6-4=2\end{array}$

Now height is

  $\begin{array}{l}height=3x+2\\ height=3(6)+2=20\end{array}$

Hence, the dimensions are $6\times 2\times 20$ .