Write and solve a polynomial equation for the situation described.
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 5 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 6 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 1920 cubic feet?

Respuesta :

Answer:

  • 2x(x-10)(x-6) = 1920 . . . the polynomial equation
  • x = 16 . . . . . . . . . . . . . . . the solution
  • dimensions: 16 ft × 32 ft × 10 ft high

Step-by-step explanation:

We can let x represent the width of the barn in feet. Then the length of the barn is 2x. The portion of width not used for stalls is (x-10). The height of that is (x-6).

So, the product of length, width not used for stalls, and height must be 1920 ft³. The equation for that is ...

... 2x(x -10)(x -6) = 1920 . . . . the equation

Solution

I generally find it convenient to use a graphing calculator to find the solutions to higher-order equations. Rearranging this so we're looking for a zero, it becomes ...

... 2x(x -10)(x -6) -1920 = 0

The graph shows one real root, at x = 16.

The dimensions of the lower floor should be 16 feet wide, 32 feet long, and a ceiling height of 10 feet.

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Alternate solution

Dividing by 2, we have

... x(x-6)(x-10) = 960

If we look at the factors of 960, we find that it is the product of 16, 10, and 6, which would let us assign x=16.

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There are 28 divisors of 960. According to the Rational Root theorem, any of them might be a solution. There are arguments that can be made regarding reasonable dimensions that suggest x will be more than 12 (6 ft ceiling, 2 ft aisle). So, the number of trial-and-error possibilities is reduced somewhat.

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