Question

A rectangular two-story horse barn is being designed for a farm. The upper floor will be used for storing hay, and the lower floor will have horse stalls that extend 6 feet from both of the longer walls. The barn's length is twice the barn's width, and the lower floor's ceiling height is 8 feet less than the barn's width. What should the dimensions of the lower floor be if the space not used for stalls is to have a volume of 3,840 cubic feet?

Answer 1

Let the width of the barn be = x feet

So length of the barn = 2x

Height of the barn = x-8

As the stalls are 6 feet longer from both ends hence, we have to find the area with width as x-12 feet

Volume of the space is = 3840 cubic feet

Hence, equation is :

$(x-12)(2x)(x-8)=3840$

Solving this we get

$2x^{3}-40x^{2}+192x=3840$

$2x^{3}-40x^{2}+192x-3840=0$

$2(x-20)(x^{2}+96)=0$

This gives x=20 and x= ±$4\sqrt{6}i$

Hence, neglecting the square root value we get x = 20 feet

Hence, the width is = x-12 = 20 - 12 = 8 feet

Length is = 2x = 2*20 = 40 feet

Height is = x-8 = 20 - 8 =12 feet

And we can cross check this by multiplying all the three dimensions to get 3840 cubic feet

$8\times40\times12=3840$ cubic feet.