Dimensions of a closed rectangular box

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The dimensions of a closed rectangular box are 70 cm, 50 cm, and 100 cm, with a maximum measurement error of 0.2 cm. The discussion focuses on estimating the maximum error in calculating the surface area, using differentials. The incorrect approach initially used volume instead of surface area, leading to confusion in unit conversion. The correct formula for surface area is 2xy + 2yh + 2xh, and all dimensions must be considered in the error analysis. Ultimately, the importance of accounting for errors in all measurements is emphasized for accurate calculations.
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The dimensions of a closed rectangular box are measured as 70 centimeters, 50 centimeters, and 100 centimeters, respectively, with the error in each measurement at most .2 centimeters. Use differentials to estimate the maximum error in calculating the surface area of the box.

Answer: ________________ square centimeters

V = LWH

dv = (WH)*DL + (LH)*DW + (LW)*DH
dv = (50)(100)(.2) + (70)(100)(.2)+(70)(50)(.2) = 3100 square centimeters


but the answer is incorrect. am i missing something?
 
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The problem is asking for error in calculating the surface area but you're doing volume and mysteriously changing the units to cm^2.
 
Surface Area of a Rectangular = 2xy + 2yh + 2xh

DV = (2y)(dx) + (2h)(dy) + (2x)(dh)
(2)(50)(0.2) + (2)(100)(0.2)+(2)(70)(0.2) = 88 which is also wrong, did i miss something agian?
 
Yes, you did. The error from the xy term, for example, is \delta x y = \delta x \times y + x \times \delta y.
 
i see that your using the product rule, but since 'y'and 2 is constant, can't you just take them out and just take the derivative of 'x'?
 
Each of the 3 dimensions is, in fact, constant. The point is that there are errors in each of the 3 measurements and you can't selectively ignore anyone of them in your error analysis.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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