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November 25, 2014

November 25, 2014

Posted by **Fabian** on Thursday, March 13, 2008 at 11:06pm.

The design of a new airplane requires a gasoline tank of constant cross-sectional area in each wing. A scale drawing of a cross section is shown here. The tank must hold 5000lb of gasoline, which has a density of 42 lb/ft3. Estimate the length of the tank.

Y0 = 1.5 ft, Y1 = 1.6 ft, Y2 = 1.8 ft, Y3 = 1.9 ft, Y4 = 2.0 ft, Y5 = Y6 = 2.1 ft

And the horizontal spacing is 1 ft.

- Math Nightmare -
**drwls**, Friday, March 14, 2008 at 7:58amWithout the drawing or a better verbal description of the wind geometry, we cannot help you. The volume of the tank must be

V = (mass)/(density)

= (5000 lb)/(42 lb/ft^3) = 119.0 ft^3

The volume V equals the cross sectional area times the length in this case. Figure out the area and use it to calculate the length

- Math Nightmare -
**katy**, Thursday, November 29, 2012 at 12:50amuse the trapezoidal rule with the scale so its

height =1, so

(1/2) (1.5+ 2*1.6+ 2*1.8+ 2*1.9+ 2*2.0+ 2*2.1+ 2.1)

which equals 11.2

so 119/11.2= 10.625 or 10.63

and that is the answer

- Math Nightmare -
**Joseph President**, Saturday, November 15, 2014 at 2:13pmSatan

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