Question

A cardboard box without a lid is to have a volume of 5,324 cm3. Find the dimensions that minimize the amount of cardboard used. (Let x, y, and z be the dimensions of the cardboard box.)

Answer 1

To minimize the amount of cardboard used, find the dimensions of the box that will give the smallest surface area. Use the method of partial derivatives to solve for x, y, and z.

Step-by-step explanation:

Let x be the length, y be the width, and z be the height of the cardboard box. We want to minimize the surface area of the box, which is given by:

S = 2xy + 2xz + 2yz

Subject to the constraint that the volume of the box is 5,324 cm3:

xy = 2662

We can use Lagrange multipliers to solve this optimization problem. We define the Lagrangian function:

L(x, y, z, λ) = 2xy + 2xz + 2yz + λ(xy - 2662)

Taking the partial derivatives of L with respect to x, y, z, and λ, we get:

∂L/∂x = 2y + 2z + λy = 0

∂L/∂y = 2x + 2z + λx = 0

∂L/∂z = 2x + 2y = 0

∂L/∂λ = xy - 2662 = 0

Solving these four equations simultaneously, we get:

x = y = z = 14 cm

Therefore, the dimensions that minimize the amount of cardboard used are x = y = z = 14 cm.

Answer 2

The area of the box is

`A(x,y,z)=xy+2xz+2yz`

which we want to minimize subject to the constraint
`xyz=5324`.

The Lagrangian is

`L(x,y,z)=xy+2xz+2yz+\lambda(xyz-5324)`

with critical points where the partial derivatives are 0:

`L_x=y+2z+\lambda yz=0`

`L_y=x+2z+\lambda xz=0`

`L_z=2x+2y+\lambda xy=0`

`L_\lambda=xyz-5324=0`

Notice that

`L_y-L_x=(x-y)+\lambda(xz-yz)=0\implies(x-y)(1+\lambda z)=0\implies x=y\text{ or }z=-\frac1\lambda`

Substituting the latter into either
`L_x=0` or
`L_y=0` will end up suggesting that
`\lambda` is infinite, so we throw out this case.

If
`x=y`, then

`L_z=0\implies4x+\lambda x^2=0\implies x=0\text{ or }x=-\frac4\lambda`

We ignore the case where
`x=0` because that would make the volume 0. Then

`x=y=-\frac4\lambda\text{ and }L_x=0\implies-\frac4\lambda+2(1331\lambda^2)+\lambda\left(-\frac4\lambda\right)(1331\lambda^2)=0`

`\implies2662\lambda^3+4=0`

`\implies\lambda=-\frac{\sqrt[3]{2}}{11}`

so we have one critical point at

`(x,y,z)=\left(22\sqrt[3]{4},22\sqrt[3]{4},\frac{11}{2\sqrt[3]{2}}\right)\approx(34.9228,34.9228,4.3654)`

which give a minimum area of about 1829.41 sq. cm.