The volume of a rectangular box is #270# #m^3#. The width of the box is 3 meters less than the length. The height is 5 meters. What is the length of the box?

Question

Autumn Armstrong · Accepted Answer

Length of the rectangular box is #9# metres.

The volume of rectangular box is # V=270 # cubic metre. Let the length of rectangular box is #L=x # metres. Then the width of rectangular box is #W=(x-3) # metres. The height of rectangular box is #H=5 # metres. We know volume of rectangular box is #V=L*W*H# #:. 270=x * (x-3) * 5 or x*(x-3)= 270/5# # or x^2-3x=54# or #x^2-3x-54=0# or # x^2-9x+6x-54=0 or x(x-9)+6(x-9)=0 # or #(x-9)(x+6)=0 :. x=9 or x= cancel(-6) ; #. Length cannot be negative. So #x=9 :. x-3 =6# Length of the rectangular box is #9# metres. [Ans]

James Atkins · Answer

undefined Let's denote the length of the box as $ L $. Given that the width is 3 meters less than the length, the width will be $ L - 3 $. The height is given as 5 meters.

The formula for the volume of a rectangular box is $ \text{Volume} = \text{Length} \times \text{Width} \times \text{Height} $.

Substituting the given values into the formula, we get:

$$ 270 = L \times (L - 3) \times 5 $$

Now, let's solve for $ L $:

$$ 270 = 5L(L - 3) $$

$$ 270 = 5L^2 - 15L $$

$$ 5L^2 - 15L - 270 = 0 $$

Now, we have a quadratic equation. We can solve this equation using the quadratic formula:

$$ L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Where $ a = 5 $, $ b = -15 $, and $ c = -270 $.

$$ L = \frac{-(-15) \pm \sqrt{(-15)^2 - 4 \times 5 \times (-270)}}{2 \times 5} $$

$$ L = \frac{15 \pm \sqrt{225 + 5400}}{10} $$

$$ L = \frac{15 \pm \sqrt{5625}}{10} $$

$$ L = \frac{15 \pm 75}{10} $$

This gives us two potential values for $ L $:

1. $ L = \frac{15 + 75}{10} = \frac{90}{10} = 9 $
2. \( L =