Optimization

[Tueseve728]

Suppose your asked to design an open top box with a square base that has a volume of 32ft^3. You must design the box to use the least possible amount of material. What should be the dimensions of the box? (Hint 1: Draw a picture and label it with vaiables. Hint 2: Write down some equation for volume and area. Hint 3: Find the absolute minimum of one of the equations.

So far I drew a pic of the box.
V=x^3
32=x^3

But do i have to do something with 5x since there are only 5 side because the box has an open top?
Thanks for your help!

 

[stapel]

So far I drew a pic of the box.
V=x^3

What does "x" stand for? And why are you assuming that the box is a cube?

Eliz.

 

[Unco]

G'day, Tueseve.

Suppose your asked to design an open top box with a square base that has a volume of 32ft^3. You must design the box to use the least possible amount of material. What should be the dimensions of the box? (Hint 1: Draw a picture and label it with vaiables. Hint 2: Write down some equation for volume and area. Hint 3: Find the absolute minimum of one of the equations.

So far I drew a pic of the box.
V=x^3
32=x^3

But do i have to do something with 5x since there are only 5 side because the box has an open top?

A box doesn't have to be cubic. We do know the base is square, though.

We have

     *---------*         
    /         /|
   /________ / | y   Width: x; length: x; height: y
  |         |  |
  |         |  |     So V = x * x * y = (x^2)y -> We are told V=32 so 32=(x^2)y [1]
  |         |  *
  |         | / x    And, as you have said, the surface area is the sum of
  |_________|/        of the five sides' areas:
       x             A =  x*x(bottom) + x*y(front) + x*y(back) + x*y(left) + x*y(right) = x^2 + 4xy [2]   
                      
      Use [1] to write [2] in terms of one variable; differentiate [2] and set to zero to minimise A.

Check that the second derivative is positive at your values for x and y so you do have a minimum.

See how you go.

Edit: error corrected!