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dolphi86 [110]
3 years ago
10

The width of a container is 5 feet less than its height. Its length is 1 foot longer than its height. The volume of the containe

r is 216 cubic feet. How tall is the container?
Mathematics
1 answer:
juin [17]3 years ago
8 0

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

Given that volume of the container is 216 cubic feet

216 = l \times w \times h

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

Solve by factoring

(h-8)(h^2+4h+27) = 0

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

Also,

h^2+4h+27 = 0

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}

\mathrm{For\:} a=1,\:b=4,\:c=27:\quad h=\frac{-4\pm \sqrt{4^2-4\cdot \:1\cdot \:27}}{2\cdot \:1}

h = \frac{-4+\sqrt{4^2-4\cdot \:1\cdot \:27}}{2}=\frac{-4+\sqrt{92}i}{2}

Therefore, on solving we get,

h=-2+\sqrt{23}i,\:h=-2-\sqrt{23}i

<em><u>Thus solutions of "h" are:</u></em>

h = 8

h=-2+\sqrt{23}i,\:h=-2-\sqrt{23}i

"h" cannot be a imaginary value

Thus the solution is h = 8

Thus the height of container is 8 feet

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