the dimensions of a closed rectangular box are measured as 30 in, 40 in.. and 60 in.. with a maximum error of 0.2 in. in each measurement.

use differentials to estimate the maximum error in calculating the volume of the box

v = xyz

dv = yz dx + xz dy + xy dz
= 2400(.2) + 1800(.2) + 1200(.2)
= 5400(.2)
= 1080 in^3

To estimate the maximum error in calculating the volume of the box using differentials, we need to find the differential of the volume equation with respect to each side length and then calculate the maximum error using these differentials.

The volume (V) of a rectangular box is given by the equation V = lwh, where l represents the length, w represents the width, and h represents the height.

Let's calculate the partial derivatives of the volume equation with respect to each side length:

dV/dl = wh
dV/dw = lh
dV/dh = lw

Now, let's calculate the maximum error using differentials for each side length:

Maximum error in length (dl) = 0.2 in
Maximum error in width (dw) = 0.2 in
Maximum error in height (dh) = 0.2 in

Using these differential values, we can calculate the maximum error in volume (dV):

dV = (dV/dl * dl) + (dV/dw * dw) + (dV/dh * dh)
= (wh * dl) + (lh * dw) + (lw * dh)

Substituting the values we have:

dV = (wh * 0.2) + (lh * 0.2) + (lw * 0.2)
= 0.2(wh + lh + lw)

Hence, the maximum error in calculating the volume of the box is approximately 0.2 times the sum of the products of each pair of side lengths.

To estimate the maximum error in calculating the volume of the box, we will use differentials. The volume of a rectangular box can be calculated by multiplying its length, width, and height.

Given that the dimensions of the box are measured as 30 in, 40 in, and 60 in, with a maximum error of 0.2 in. in each measurement, we can determine the maximum error in each dimension as follows:

For the length:
- Original length: 30 in
- Maximum error: 0.2 in
- Relative or fractional error: 0.2/30 = 0.00667 (rounded to 5 decimal places)

For the width:
- Original width: 40 in
- Maximum error: 0.2 in
- Relative or fractional error: 0.2/40 = 0.005 (rounded to 5 decimal places)

For the height:
- Original height: 60 in
- Maximum error: 0.2 in
- Relative or fractional error: 0.2/60 = 0.00333 (rounded to 5 decimal places)

To estimate the maximum error in calculating the volume, we need to find the partial derivatives of the volume function with respect to each variable (length, width, and height). Then, we multiply these partial derivatives by their respective relative errors and sum them up.

The volume of the box is given by:
V = length * width * height

Taking the partial derivative with respect to length (L):
∂V/∂L = width * height

Taking the partial derivative with respect to width (W):
∂V/∂W = length * height

Taking the partial derivative with respect to height (H):
∂V/∂H = length * width

Next, we multiply each partial derivative by its relative error and sum them up to estimate the maximum error in volume:
ΔV = (∂V/∂L) * (ΔL) + (∂V/∂W) * (ΔW) + (∂V/∂H) * (ΔH)
ΔV = (width * height) * (ΔL) + (length * height) * (ΔW) + (length * width) * (ΔH)

Substituting the values:
ΔV = (40 * 60) * (0.00667) + (30 * 60) * (0.005) + (30 * 40) * (0.00333)

Calculating the above expression will give the estimated maximum error in the volume of the box.