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Sveta_85 [38]
4 years ago
11

A box has a volume of 192 cubic inches, a length that is twice as long as its width, and a height that is 2 inches greater than

the width. What are the dimensions of this box? Give your answer in the form l,w,h as comma separated values. For example if the box has a length of 1, a width of 2, and a height of 3, you would submit the answer 1,2,3.
Mathematics
1 answer:
Volgvan4 years ago
4 0

Answer:

Length = 8 inches

Width = 4 inches

Height = 6 inches

Step-by-step explanation:

Volume of the box = 192 cubic inches

Volume of a box = length × width × height

Let

Width = x

Length = 2x

Height = x + 2

Volume of a box = length × width × height

192 = 2x * x * (x + 2)

192 = 2x^2 (x + 2)

192 = 2x^3 + 4x^2

Divide through by 2

96 = x^3 + 2x^2

Subtract 96 from both sides

x ^3 + 2x^2 - 96 = 0

Factorise

(x - 4) (x^2 + 6x + 24) = 0

x^2 + 6x + 24 has no real x-value

So,

Divide both sides by (x^2 + 6x + 24).

0 ÷ (x^2 + 6x + 24) = 0

So,

(x - 4) = 0

x = 4

Width = x = 4 inches

Length = 2x

= 2(4)

= 8 inches

Height = x + 2

= 4 + 2

= 6 inches

Check:

Volume of a box = length × width × height

= (8 * 4 * 6) inches

= 192 inches

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Answer:

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Step-by-step explanation:

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to find the max height evaluate f(2.75 )

f(2.75 ) = - 16( 2.75)² + 88(2.75) + 12 = - 121 + 242 + 12 = 133 feet




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