The height of container is 8 feet
<em><u>Solution:</u></em>
Let "w" be the width of container
Let "l" be the length of container
Let "h" be the height of container
The width of a container is 5 feet less than its height
Therefore,
width = height - 5
w = h - 5 ------ eqn 1
Its length is 1 foot longer than its height
length = 1 + height
l = 1 + h ---------- eqn 2
<em><u>The volume of container is given as:</u></em>
![v = length \times width \times height](https://tex.z-dn.net/?f=v%20%3D%20length%20%5Ctimes%20width%20%5Ctimes%20height)
Given that volume of the container is 216 cubic feet
![216 = l \times w \times h](https://tex.z-dn.net/?f=216%20%3D%20l%20%5Ctimes%20w%20%5Ctimes%20h)
Substitute eqn 1 and eqn 2 in above formula
![216 = (1 + h) \times (h-5) \times h\\\\216 = (h+h^2)(h-5)\\\\216 = h^2-5h+h^3-5h^2\\\\216 = h^3-4h^2-5h\\\\h^3-4h^2-5h-216 = 0](https://tex.z-dn.net/?f=216%20%3D%20%281%20%2B%20h%29%20%5Ctimes%20%28h-5%29%20%5Ctimes%20h%5C%5C%5C%5C216%20%3D%20%28h%2Bh%5E2%29%28h-5%29%5C%5C%5C%5C216%20%3D%20h%5E2-5h%2Bh%5E3-5h%5E2%5C%5C%5C%5C216%20%3D%20h%5E3-4h%5E2-5h%5C%5C%5C%5Ch%5E3-4h%5E2-5h-216%20%3D%200)
Solve by factoring
![(h-8)(h^2+4h+27) = 0](https://tex.z-dn.net/?f=%28h-8%29%28h%5E2%2B4h%2B27%29%20%3D%200)
Use the zero factor principle
If ab = 0 then a = 0 or b = 0 ( or both a = 0 and b = 0)
Therefore,
![h - 8 = 0\\\\h = 8](https://tex.z-dn.net/?f=h%20-%208%20%3D%200%5C%5C%5C%5Ch%20%3D%208)
Also,
![h^2+4h+27 = 0](https://tex.z-dn.net/?f=h%5E2%2B4h%2B27%20%3D%200)
Solve by quadratic equation formula
![\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}](https://tex.z-dn.net/?f=%5Ctext%20%7BFor%20a%20quadratic%20equation%20%7D%20a%20x%5E%7B2%7D%2Bb%20x%2Bc%3D0%2C%20%5Ctext%20%7B%20where%20%7D%20a%20%5Cneq%200%5C%5C%5C%5Cx%3D%5Cfrac%7B-b%20%5Cpm%20%5Csqrt%7Bb%5E%7B2%7D-4%20a%20c%7D%7D%7B2%20a%7D)
![\mathrm{For\:} a=1,\:b=4,\:c=27:\quad h=\frac{-4\pm \sqrt{4^2-4\cdot \:1\cdot \:27}}{2\cdot \:1}](https://tex.z-dn.net/?f=%5Cmathrm%7BFor%5C%3A%7D%20a%3D1%2C%5C%3Ab%3D4%2C%5C%3Ac%3D27%3A%5Cquad%20h%3D%5Cfrac%7B-4%5Cpm%20%5Csqrt%7B4%5E2-4%5Ccdot%20%5C%3A1%5Ccdot%20%5C%3A27%7D%7D%7B2%5Ccdot%20%5C%3A1%7D)
![h = \frac{-4+\sqrt{4^2-4\cdot \:1\cdot \:27}}{2}=\frac{-4+\sqrt{92}i}{2}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B-4%2B%5Csqrt%7B4%5E2-4%5Ccdot%20%5C%3A1%5Ccdot%20%5C%3A27%7D%7D%7B2%7D%3D%5Cfrac%7B-4%2B%5Csqrt%7B92%7Di%7D%7B2%7D)
Therefore, on solving we get,
![h=-2+\sqrt{23}i,\:h=-2-\sqrt{23}i](https://tex.z-dn.net/?f=h%3D-2%2B%5Csqrt%7B23%7Di%2C%5C%3Ah%3D-2-%5Csqrt%7B23%7Di)
<em><u>Thus solutions of "h" are:</u></em>
h = 8
![h=-2+\sqrt{23}i,\:h=-2-\sqrt{23}i](https://tex.z-dn.net/?f=h%3D-2%2B%5Csqrt%7B23%7Di%2C%5C%3Ah%3D-2-%5Csqrt%7B23%7Di)
"h" cannot be a imaginary value
Thus the solution is h = 8
Thus the height of container is 8 feet