Question

The dimensions of a closed rectangular box are measured as 80 cm, 61 cm, and 29 cm, respectively, with a possible error of 0.2 cm in each dimension. Use differentials to estimate the maximum error in calculating the surface area of the box. (Round your answer to one decimal place.)

Answer 1

Answer:

$136cm^2$

Step-by-step explanation:

We are given:

x = 80cm

y = 61cm

z = 29cm

Let's say,

f(x,y,z) = 2xy + 2xz + 2yz.

Therefore,

(x,y,z) = (80, 61, 29)

and Δx = Δy = Δz = 0.2

Making use of partial derivatives, we have:

$f_x = 2y + 2z$

$f_y = 2x + 2z$

$f_z = 2x + 2y$

Therefore, to solve fx, fy and fz we have:

f_x (80, 61, 29)

= (2*61)+(2*29) =180

f_y (80, 61, 29)

= (2*80)+(2*29) = 218

f_z (80, 61, 29)

= (2*80)+(2*61)=282

dz at (80, 61, 29) =

dz = 180dx + 218dy + 282dz.

Δz = 180Δx + 218Δy + 282Δz

=(180 + 218 + 282) * 0.2

= 680 * 0.2

= 136

Therefore, the max error is 136sq.cm

Answer 2

The maximum error in calculating surface area is $136 cm^{2}$

Surface area :

The maximum error in calculating surface area is given as,

                      $\Delta A = 4d(x+y+z)$

where x , y and z are dimensions of rectangular box.

Given that,  $x=80,y=61,z=29,d=0.2$

Substitute values in above relation.

              $\Delta A = 4*0.2(80+61+29)\\\\\Delta A = 0.8*170\\\\\Delta A = 136cm^{2}$

The maximum error in calculating surface area is $136 cm^{2}$

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