# Determine the dimensions of a rectangular box, open at the top

• naspek
In summary, the problem asks for the dimensions of a rectangular box having a volume of 4 m3, and requiring the least amount of material possible for its construction. You are to find the dimensions of the box when y=0, z=0 using the second partials test.
naspek

## Homework Statement

Determine the dimensions of a rectangular box, open at the top, having volume
4 m3, and requiring the least amount of material for its construction. Use the
second partials test. (Hint: Take advantage of the symmetry of the problem.)

## The Attempt at a Solution

Volume, V= 4m^3
let x = length
y = width
z = height

4m^3 = xyz
x = 4/yz

because it is an open at the top rectangular box,
the Surface, S = 2xz + 2yz + xy

substitute x=4/yz inside the surface equation..
S = 8/y + 2yz + 4/z

to find the critical points, take the partial with respect to y and z.. the equal it to zero..

S'(y)= 2z - 8/y^2
S'(z)= 2y - 4/z^2

solve the equations, i get,
when y = 0, z = 0..
y = 8, z = 1/16

the problem is, what should i do next?

When y = 0, 8/y^2 is undefined! When I set Sy and Sz to 0, I get different values than you did.

Also, you're supposed to use the second partials test.

ic... ok..

so..

S'(y) = 2z - 8/y^2
S'(yy) = 16/y^3
S'(yz) = 2

S'(z) = 2y - 4/z^2
S'(zz) = 8/z^3
S'(zy) = 2

what should i do next?

You still haven't found the values of y and z for which Sy = 0 and Sz = 0.

Also, you haven't used the second partials test; all you have done is calculate all the second partials.

ok.. for.. S'(y)=0 -----> 2z - 8/y^2=0

...for.. S'(z)=0 -----> 2y - 4/z^2 =

ok.. by letting z=4/y^2, i get y = 0, +8^1/2, -8^1/2

am i got it correct now?

is it possible if i use lagrange multiplier method?

done with the equation for x, y and z..
got some miscalculation just now..
i got y=2, z= 1, x = 2..
why i should do the second partial test?

naspek said:
done with the equation for x, y and z..
got some miscalculation just now..
i got y=2, z= 1, x = 2..
why i should do the second partial test?
Those are the values I got.

You should use the second partial test for two reasons:
1. The problem asks you to use it.
2. So you can tell whether these values give you a maximum value or a minimum value of S.

thanks.. u are really helpful =)

## 1. How do you measure the dimensions of a rectangular box?

To determine the dimensions of a rectangular box, you will need a ruler or measuring tape. Measure the length, width, and height of the box and record the measurements in the same unit of measurement.

## 2. What is the formula for calculating the volume of a rectangular box?

The formula for calculating the volume of a rectangular box is length x width x height. This will give you the total amount of space inside the box in cubic units.

## 3. Can you determine the dimensions of a rectangular box if it is not open at the top?

Yes, you can still determine the dimensions of a rectangular box even if it is not open at the top. You will need to measure the length, width, and height of the box from the outside, using a ruler or measuring tape.

## 4. How do you find the length, width, and height if the rectangular box is not a perfect shape?

If the rectangular box is not a perfect shape, you can still measure the length, width, and height using the same method as before. However, you may need to break down the measurements into smaller sections and add them together to get the total length, width, and height.

## 5. Is it necessary to measure all three dimensions of a rectangular box or can I just measure two?

In order to accurately determine the dimensions of a rectangular box, it is necessary to measure all three dimensions - length, width, and height. If you only measure two dimensions, you will not have the complete picture and may not get an accurate measurement of the box's volume.

• Calculus and Beyond Homework Help
Replies
1
Views
614
• Calculus and Beyond Homework Help
Replies
5
Views
2K
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
679
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
752
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
45
Views
4K
• Calculus and Beyond Homework Help
Replies
6
Views
935