SOLUTION: A Metal Sleeve Of Length 20 Cm Has Rectangular Cross-Section 10 Cm By 8 Cm. The Metal Has Uniform Thickness, Xcm, Along The Sleeve, And The Total Volume Of metal in the sleeve is 4

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A Metal Sleeve Of Length 20 Cm Has Rectangular Cross-Section 10 Cm By 8 Cm. The Metal Has Uniform Thickness, Xcm, Along The Sleeve, And The Total Volume Of metal in the sleeve is 4      Log On


   



Question 1203201: A Metal Sleeve Of Length 20 Cm Has Rectangular Cross-Section 10 Cm By 8 Cm. The Metal Has Uniform Thickness, Xcm, Along The Sleeve, And The Total Volume Of metal in the sleeve is 495cm3. Derive the equation 16x^2-144x+99=0.

Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
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A Metal Sleeve Of Length 20 Cm Has Rectangular Cross-Section 10 Cm By 8 Cm (highlight%28outer_dimensions%29).
The Metal Has Uniform Thickness, x cm, Along The Sleeve, And The Total Volume Of metal in the sleeve is 495 cm3.
Derive the equation 16x^2-144x+99=0.
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The outer dimensions of the sleeve are 10 cm by 8 cm.
The metallic walls have thickness of x cm, uniformly.
So, the empty space in cross-section has dimensions (10-2x) cm by (8-2x) cm.
We subtract 2x to account for the thickness of the two walls in each direction.


Therefore, the area of the metal in cross-section is

    area = 10*8 - (10-2x)*(8-2x) cm^2.


    +---------------------------------------------------------+
    |    Since the length is 20 cm, the volume of the metal   |
    |        is 20 times the area of the cross-section.       |
    +---------------------------------------------------------+


So, your equation for x is

    -20*(4x^2 - 36x) = 495  cm^3.


Divide both sides by 5 and simplify further step by step.

Nice exercise for you to complete it on your own.

You may report me about your progress.

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